Skip to main content

Full text of "Vector Calculus"

See other formats


Vector Calculus 



Fourth Edition 



if* Jr -p 





Susan Jane Colley 



This page intentionally left blank 



Vector Calculus 



This page intentionally left blank 



Vector Calculus! 4™ 



Susan Jane Colley 

Oberlin College 



PEARSON 



Boston Columbus Indianapolis New York San Francisco Upper Saddle River 
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto 
Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo 



Editor in Chief: Deirdre Lynch 

Senior Acquisitions Editor: William Hoffman 

Sponsoring Editor: Caroline Celano 

Editorial Assistant: Brandon Rawnsley 

Senior Managing Editor: Karen Wernholm 

Production Project Manager: Beth Houston 

Executive Marketing Manager: Jeff Weidenaar 

Marketing Assistant: Caitlin Crain 

Senior Author Support/Technology Specialist: Joe Vetere 

Rights and Permissions Advisor: Michael Joyce 

Manufacturing Buyer: Debbie Rossi 

Design Manager: Andrea Nix 

Senior Designer: Beth Paquin 

Production Coordination and Composition: Aptara, Inc. 

Cover Designer: Suzanne Duda 

Cover Image: Alessandro Delia Bella/AP Images 

Many of the designations used by manufacturers and sellers to distinguish their products are claimed as 
trademarks. Where those designations appear in this book, and Pearson Education was aware of a trademark 
claim, the designations have been printed in initial caps or all caps. 

Library of Congress Cataloging-in-Publication Data 

Colley, Susan Jane. 

Vector calculus / Susan Jane Colley. - 4th ed. 
p. cm. 
Includes index. 

ISBN-13: 978-0-321-78065-2 

ISBN-10: 0-321-78065-5 
1. Vector analysis. I. Title. 
QA433.C635 2012 
515'.63-dc23 

2011022433 

Copyright © 2012, 2006, 2002 Pearson Education, Inc. 

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any 
means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in 
the United States of America. For information on obtaining permission for use of material in this work, please submit a written request 
to Pearson Education, Inc., Rights and Contracts Department, 501 Boylston Street, Suite 900, Boston, MA 021 16, fax your request to 
617-671-3447, or e-mail at http://www.pearsoned.com/legal/permissions.htm. 

123456789 10— EB— 15 14 13 12 11 



PEARSON 



www.pearsonhighered.com 



ISBN 13: 978-0-321-78065-2 
ISBN 10: 0-321-78065-5 



To Will and Diane, 

with love 



About the Author 
Susan Jane Colley 




Susan Colley is the Andrew and Pauline Delaney Professor of Mathematics at 
Oberlin College and currently Chair of the Department, having also previously 
served as Chair. 

She received S.B. and Ph.D. degrees in mathematics from the Massachusetts 
Institute of Technology prior to joining the faculty at Oberlin in 1983. 

Her research focuses on enumerative problems in algebraic geometry, partic- 
ularly concerning multiple-point singularities and higher-order contact of plane 
curves. 

Professor Colley has published papers on algebraic geometry and commuta- 
tive algebra, as well as articles on other mathematical subjects. She has lectured 
internationally on her research and has taught a wide range of subjects in under- 
graduate mathematics. 

Professor Colley is a member of several professional and honorary societies, 
including the American Mathematical Society, the Mathematical Association of 
America, Phi Beta Kappa, and Sigma Xi. 



Contents 



Preface ix 
To the Student: Some Preliminary Notation xv 



1 Vectors 1 

1.1 Vectors in Two and Three Dimensions 1 

1.2 More About Vectors 8 

1.3 The Dot Product 18 

1.4 The Cross Product 27 

1.5 Equations for Planes; Distance Problems 40 

1.6 Some n-dimensional Geometry 48 

1.7 New Coordinate Systems 62 
True/False Exercises for Chapter 1 75 
Miscellaneous Exercises for Chapter 1 75 



Differentiation in Several Variables 82 

2.1 Functions of Several Variables; Graphing Surfaces 82 

2.2 Limits 97 

2.3 The Derivative 116 

2.4 Properties; Higher-order Partial Derivatives 134 

2.5 The Chain Rule 142 

2.6 Directional Derivatives and the Gradient 158 

2.7 Newton's Method (optional) 176 
True/False Exercises for Chapter 2 182 
Miscellaneous Exercises for Chapter 2 183 



3 Vector-Valued Functions 189 

3.1 Parametrized Curves and Kepler's Laws 189 

3.2 Arclength and Differential Geometry 202 

3.3 Vector Fields: An Introduction 221 

3.4 Gradient, Divergence, Curl, and the Del Operator 227 
True/False Exercises for Chapter 3 237 
Miscellaneous Exercises for Chapter 3 237 



4 Maxima and Minima in Several Variables 244 

4.1 Differentials and Taylor's Theorem 244 

4.2 Extrema of Functions 263 

4.3 Lagrange Multipliers 278 

4.4 Some Applications of Extrema 293 
True/False Exercises for Chapter 4 305 
Miscellaneous Exercises for Chapter 4 306 



Multiple Integration 310 

5.1 Introduction: Areas and Volumes 310 

5.2 Double Integrals 314 

5.3 Changing the Order of Integration 334 

5.4 Triple Integrals 337 

5.5 Change of Variables 349 

5.6 Applications of Integration 373 

5.7 Numerical Approximations of Multiple Integrals (optional) 388 
True/False Exercises for Chapter 5 401 
Miscellaneous Exercises for Chapter 5 403 



6 Line Integrals 408 

6.1 Scalar and Vector Line Integrals 408 

6.2 Green's Theorem 429 

6.3 Conservative Vector Fields 439 
True/False Exercises for Chapter 6 450 
Miscellaneous Exercises for Chapter 6 451 



7 Surface Integrals and Vector Analysis 455 

7.1 Parametrized Surfaces 455 

7.2 Surface Integrals 469 

7.3 Stokes's and Gauss's Theorems 490 

7.4 Further Vector Analysis; Maxwell's Equations 510 
True/False Exercises for Chapter 7 522 
Miscellaneous Exercises for Chapter 7 523 



8 Vector Analysis in Higher Dimensions 530 

8.1 An Introduction to Differential Forms 530 

8.2 Manifolds and Integrals of k-forms 536 

8.3 The Generalized Stokes's Theorem 553 
True/False Exercises for Chapter 8 561 
Miscellaneous Exercises for Chapter 8 561 

Suggestions for Further Reading 563 

Answers to Selected Exercises 565 

Index 599 



Preface 



Physical and natural phenomena depend on a complex array of factors. The sociol- 
ogist or psychologist who studies group behavior, the economist who endeavors 
to understand the vagaries of a nation's employment cycles, the physicist who 
observes the trajectory of a particle or planet, or indeed anyone who seeks to 
understand geometry in two, three, or more dimensions recognizes the need to 
analyze changing quantities that depend on more than a single variable. Vec- 
tor calculus is the essential mathematical tool for such analysis. Moreover, it 
is an exciting and beautiful subject in its own right, a true adventure in many 
dimensions. 

The only technical prerequisite for this text, which is intended for a 
sophomore-level course in multivariable calculus, is a standard course in the cal- 
culus of functions of one variable. In particular, the necessary matrix arithmetic 
and algebra (not linear algebra) are developed as needed. Although the mathe- 
matical background assumed is not exceptional, the reader will still be challenged 
in places. 

My own objectives in writing the book are simple ones: to develop in students 
a sound conceptual grasp of vector calculus and to help them begin the transition 
from first-year calculus to more advanced technical mathematics. I maintain that 
the first goal can be met, at least in part, through the use of vector and matrix 
notation, so that many results, especially those of differential calculus, can be 
stated with reasonable levels of clarity and generality. Properly described, results 
in the calculus of several variables can look quite similar to those of the calculus 
of one variable. Reasoning by analogy will thus be an important pedagogical tool. 
I also believe that a conceptual understanding of mathematics can be obtained 
through the development of a good geometric intuition. Although I state many 
results in the case of n variables (where n is arbitrary), I recognize that the most 
important and motivational examples usually arise for functions of two and three 
variables, so these concrete and visual situations are emphasized to explicate the 
general theory. Vector calculus is in many ways an ideal subject for students 
to begin exploration of the interrelations among analysis, geometry, and matrix 
algebra. 

Multivariable calculus, for many students, represents the beginning of signif- 
icant mathematical maturation. Consequently, I have written a rather expansive 
text so that they can see that there is a story behind the results, techniques, and 
examples — that the subject coheres and that this coherence is important for prob- 
lem solving. To indicate some of the power of the methods introduced, a number 
of topics, not always discussed very fully in a first multivariable calculus course, 
are treated here in some detail: 

• an early introduction of cylindrical and spherical coordinates (§1.7); 

• the use of vector techniques to derive Kepler's laws of planetary motion 
(§3.1); 

• the elementary differential geometry of curves in R 3 , including discussion 
of curvature, torsion, and the Frenet-Serret formulas for the moving frame 
(§3.2); 

• Taylor's formula for functions of several variables (§4.1); 



X Preface 

• the use of the Hessian matrix to determine the nature (as local extrema) of 
critical points of functions of n variables (§4.2 and §4.3); 

• an extended discussion of the change of variables formula in double and triple 
integrals (§5.5); 

• applications of vector analysis to physics (§7.4); 

• an introduction to differential forms and the generalized Stokes's theorem 
(Chapter 8). 

Included are a number of proofs of important results. The more techni- 
cal proofs are collected as addenda at the ends of the appropriate sections so 
as not to disrupt the main conceptual flow and to allow for greater flexibility 
of use by the instructor and student. Nonetheless, some proofs (or sketches of 
proofs) embody such central ideas that they are included in the main body of the 
text. 

New in the Fourth Edition 

I have retained the overall structure and tone of prior editions. New features in 
this edition include the following: 

• 210 additional exercises, at all levels; 

• a new, optional section (§5.7) on numerical methods for approximating 
multiple integrals; 

• reorganization of the material on Newton's method for approximating 
solutions to systems of n equations in n unknowns to its own (optional) 
section (§2.7); 

• new proofs in Chapter 2 of limit properties (in §2.2) and of the general 
multivariable chain rule (Theorem 5.3 in §2.5); 

• proofs of both single-variable and multivariable versions of Taylor's theorem 
in §4.1; 

• various additional refinements and clarifications throughout the text, 
including many new and revised examples and explanations; 

• new Microsoft® PowerPoint® files and Wolfram Mathematica® notebooks 
that coordinate with the text and that instructors may use in their teaching 
(see "Ancillary Materials" below). 

How to Use This Book 

There is more material in this book than can be covered comfortably during a single 
semester. Hence, the instructor will wish to eliminate some topics or subtopics — or 
to abbreviate the rather leisurely presentations of limits and differentiability. Since 
I frequently find myself without the time to treat surface integrals in detail, I have 
separated all material concerning parametrized surfaces, surface integrals, and 
Stokes's and Gauss's theorems (Chapter 7), from that concerning line integrals 
and Green's theorem (Chapter 6). In particular, in a one-semester course for 
students having little or no experience with vectors or matrices, instructors can 
probably expect to cover most of the material in Chapters 1-6, although no doubt 
it will be necessary to omit some of the optional subsections and to downplay 



Preface xi 



many of the proofs of results. A rough outline for such a course, allowing for 
some instructor discretion, could be the following: 

Chapter 1 8-9 lectures 



If students have a richer background (so that much of the material in Chapter 1 
can be left largely to them to read on their own), then it should be possible to treat 
a good portion of Chapter 7 as well. For a two-quarter or two-semester course, 
it should be possible to work through the entire book with reasonable care and 
rigor, although coverage of Chapter 8 should depend on students' exposure to 
introductory linear algebra, as somewhat more sophistication is assumed there. 

The exercises vary from relatively routine computations to more challenging 
and provocative problems, generally (but not invariably) increasing in difficulty 
within each section. In a number of instances, groups of problems serve to intro- 
duce supplementary topics or new applications. Each chapter concludes with a 
set of miscellaneous exercises that both review and extend the ideas introduced 
in the chapter. 

A word about the use of technology. The text was written without reference 
to any particular computer software or graphing calculator. Most of the exercises 
can be solved by hand, although there is no reason not to turn over some of the 
more tedious calculations to a computer. Those exercises that require a computer 
for computational or graphical purposes are marked with the symbol ^ and 
should be amenable to software such as Mathematical ', Maple®, or MATLAB. 

Ancillary Materials 

In addition to this text a Student Solutions Manual is available. An Instructor's 
Solutions Manual, containing complete solutions to all of the exercises, is 
available to course instructors from the Pearson Instructor Resource Center 
(www.pearsonhighered.com/irc), as are many Microsoft® PowerPoint® files and 
Wolfram Mathematica® notebooks that can be adapted for classroom use. The 
reader can find errata for the text and accompanying solutions manuals at the 
following address: 

www.oberlin.edu/math/faculty/colley/VCErrata.html 
Acknowledgments 

I am very grateful to many individuals for sharing with me their thoughts and ideas 
about multivariable calculus. I would like to express particular appreciation to my 
Oberlin colleagues (past and present) Bob Geitz, Kevin Hartshorn, Michael Henle 
(who, among other things, carefully read the draft of Chapter 8), Gary Kennedy, 
Dan King, Greg Quenell, Michael Raney, Daniel Steinberg, Daniel Styer, Richard 
Vale, Jim Walsh, and Elizabeth Wilmer for their conversations with me. I am also 
grateful to John Alongi, Northwestern University; Matthew Conner, University 
of California, Davis; Henry C. King, University of Maryland; Stephen B. Maurer, 



Chapter 2 
Chapter 3 
Chapter 4 
Chapter 5 
Chapter 6 



9 lectures 

4- 5 lectures 

5- 6 lectures 



8 lectures 
4 lectures 



38^11 lectures 



Swarthmore College; Karen Saxe, Macalester College; David Singer, Case West- 
ern Reserve University; and Mark R. Treuden, University of Wisconsin at Stevens 
Point, for their helpful comments. Several colleagues reviewed various versions 
of the manuscript, and I am happy to acknowledge their efforts and many fine 
suggestions. In particular, for the first three editions, I thank the following re- 
viewers: 

Raymond J. Cannon, Baylor University; 
Richard D. Carmichael, Wake Forest University; 
Stanley Chang, Wellesley College; 

Marcel A. F. Deruaz, University of Ottawa (now emeritus); 

Krzysztof Galicki, University of New Mexico (deceased); 

Dmitry Gokhman, University of Texas at San Antonio; 

Isom H. Herron, Rensselaer Polytechnic Institute; 

Ashwani K. Kapila, Rensselaer Polytechnic Institute; 

Christopher C. Leary, State University of New York, College at Geneseo; 

David C. Minda, University of Cincinnati; 

Jeffrey Morgan, University of Houston; 

Monika Nitsche, University of New Mexico; 

Jeffrey L. Nunemacher, Ohio Wesleyan University; 

Gabriel Prajitura, State University of New York, College at Brockport; 

Florin Pop, Wagner College; 

John T. Scheick, The Ohio State University (now emeritus); 

Mark Schwartz, Ohio Wesleyan University; 

Leonard M. Smiley, University of Alaska, Anchorage; 

Theodore B. Stanford, New Mexico State University; 

James Stasheff, University of North Carolina at Chapel Hill (now emeritus); 

Saleem Watson,California State University, Long Beach; 

Floyd L. Williams, University of Massachusetts, Amherst (now emeritus). 

For the fourth edition, I thank: 

Justin Corvino, Lafayette College; 

Carrie Finch, Washington and Lee University; 

Soomin Kim, Johns Hopkins University; 

Tanya Leise, Amherst College; 

Bryan Mosher, University of Minnesota. 

Many people at Oberlin College have been of invaluable assistance through- 
out the production of all the editions of Vector Calculus. I would especially like 
to thank Ben Miller for his hard work establishing the format for the initial drafts 
and Stephen Kasperick-Postellon for his manifold contributions to the typeset- 
ting, indexing, proofreading, and friendly critiquing of the original manuscript. I 
am very grateful to Linda Miller and Michael Bastedo for their numerous typo- 
graphical contributions and to Catherine Murillo for her help with any number of 
tasks. Thanks also go to Joshua Davis and Joaquin Espinoza Goodman for their 
assistance with proofreading. Without the efforts of these individuals, this project 
might never have come to fruition. 

The various editorial and production staff members have been most kind and 
helpful to me. For the first three editions, I would like to express my appreciation 
to my editor, George Lobell, and his editorial assistants Gale Epps, Melanie 
Van Benthuysen, and Jennifer Urban; to production editors Nicholas Romanelli, 
Barbara Mack, and Debbie Ryan at Prentice Hall, and Lori Hazzard at Interactive 
Composition Corporation; to Ron Weickart and the staff at Network Graphics 



Preface xiii 



for their fine rendering of the figures, and to Tom Benfatti of Prentice Hall for 
additional efforts with the figures; and to Dennis Kletzing for his careful and 
enthusiastic composition work. For this edition, it is a pleasure to acknowledge 
my upbeat editor, Caroline Celano, and her assistant, Brandon Rawnsley; they 
have made this new edition fun to do. In addition, I am most grateful to Beth 
Houston, my production manager at Pearson, Jogender Taneja and the staff at 
Aptara, Inc., Donna Mulder, Roger Lipsett, and Thomas Wegleitner. 

Finally, I thank the many Oberlin students who had the patience to listen to 
me lecture and who inspired me to write and improve this volume. 

SJC 

sjcolley@math.oberlin.edu 



This page intentionally left blank 



To the Student: 
Some Preliminary 

Notation 

Here are the ideas that you need to keep in mind as you read this book and learn 
vector calculus. 

Given two sets A and B, I assume that you are familiar with the notation 
A U B for the union of A and B — those elements that are in either A or B (or 
both): 

AU B = {x\x € Aorx € B}. 

Similarly, A n B is used to denote the intersection of A and B — those elements 
that are in both A and B : 

A(~)B = {x\x<=A andx e B}. 

The notation A c B, or A c B, indicates that A is a subset of B (possibly empty 
or equal to B). 

One-dimensional space (also called the real line or R) is just a straight line. 
We put real number coordinates on this line by placing negative numbers on the 
left and positive numbers on the right. (See Figure 1.) 

Two-dimensional space, denoted R 2 , is the familiar Cartesian plane. If we 
construct two perpendicular lines (the x- and y-coordinate axes), set the origin 
as the point of intersection of the axes, and establish numerical scales on these 
lines, then we may locate a point in R 2 by giving an ordered pair of numbers (x , y), 
the coordinates of the point. Note that the coordinate axes divide the plane into 
four quadrants. (See Figure 2.) 

Three-dimensional space, denoted R 3 , requires three mutually perpendicular 
coordinate axes (called the x-, y- and z-axes) that meet in a single point (called 
the origin) in order to locate an arbitrary point. Analogous to the case of R 2 , if we 
establish scales on the axes, then we can locate a point in R 3 by giving an ordered 
triple of numbers (x, y, z). The coordinate axes divide three-dimensional space 
into eight octants. It takes some practice to get your sense of perspective correct 
when sketching points in R 3 . (See Figure 3.) Sometimes we draw the coordinate 
axes in R 3 in different orientations in order to get a better view of things. However, 
we always maintain the axes in a right-handed configuration. This means that 
if you curl the fingers of your right hand from the positive x-axis to the positive 
y-axis, then your thumb will point along the positive z-axis. (See Figure 4.) 

Although you need to recall particular techniques and methods from the 
calculus you have already learned here are some of the more important concepts 
to keep in mind: Given a function f(x), the derivative f'(x) is the limit (if it exists) 
of the difference quotient of the function: 

f (x ) = hm . 

J v ft^o h 



XVI To the Student: Some Preliminary Notation 



(2,4,5) 








Figure 3 Three-dimensional 
space R 3 . Selected points are 
graphed. 



Figure 4 The x-, y-, and z-axes in R 3 are always 
drawn in a right-handed configuration. 



The significance of the derivative /'(xo) is that it measures the slope of the line 
tangent to the graph of / at the point (xo, /(xo)). (See Figure 5.) The derivative 
may also be considered to give the instantaneous rate of change of / at x = Xq. 
We also denote the derivative f'(x) by df/dx. 

The definite integral f b f(x) dx of / on the closed interval [a , b] is the limit 
(provided it exists) of the so-called Riemann sums of /: 



Figure 5 The derivative f'(xo) is 
the slope of the tangent line to 
y = f (x) at (x 0 , /(x 0 )). 



f 



f(x)dx = lim 

all Ax,-*0 



£/(x*)Ax,. 



(=1 



Here a = xo < x\ < xi < • • • < x„ = b denotes a partition of [a, b] into subin- 
tervals [x ( _i, x, ], the symbol A (the length of the subinterval), and 

x* denotes any point in [x/_i , x,-]. If f(x) > 0 on [a, b], then each term f(x*)Axj 
in the Riemann sum is the area of a rectangle related to the graph of /. The 
Riemann sum YH=i f( x *)^- x < mus approximates the total area under the graph 
of / between x = a and x = b. (See Figure 6.) 




Figure 6 If f(x) > 0 on [a, b], then the Riemann sum 
approximates the area under y = f(x) by giving the sum 
of areas of rectangles. 



To the Student: Some Preliminary Notation XVII 




The definite integral f b f(x)dx, if it exists, is taken to represent the area 
under y = f(x) between x = a and x = b. (See Figure 7.) 

The derivative and the definite integral are connected by an elegant result 
known as the fundamental theorem of calculus. Let f(x) be a continuous func- 
tion of one variable, and let F(x) be such that F'(x) = f(x). (The function F is 
called an antiderivative of /.) Then 



I 



b 

f(x)dx = F(b)-F(a); 



d f x 

2. — / f(t)dt = f{x). 

dx J a 



Finally, the end of an example is denoted by the symbol ♦ and the end of a 
proof by the symbol ■. 



This page intentionally left blank 



Vector Calculus 



This page intentionally left blank 



Vectors 



1.1 Vectors in Two and Three 
Dimensions 

1 .2 More About Vectors 

1 .3 The Dot Product 

1 .4 The Cross Product 

1.5 Equations for Planes; 
Distance Problems 

1 .6 Some n-dimensional 
Geometry 

1.7 New Coordinate Systems 

True/False Exercises for 
Chapter 1 

Miscellaneous Exercises for 
Chapter 1 



1 .1 Vectors in Two and Three Dimensions 

For your study of the calculus of several variables, the notion of a vector is 
fundamental. As is the case for many of the concepts we shall explore, there are 
both algebraic and geometric points of view. You should become comfortable 
with both perspectives in order to solve problems effectively and to build on your 
basic understanding of the subject. 

Vectors in R 2 and R 3 : The Algebraic Notion 



DEFINITION 1.1 A vector in R 2 is simply an ordered pair of real numbers. 
That is, a vector in R 2 may be written as 

(a u a 2 ) (e.g., (1,2) or (n, 17)). 

Similarly, a vector in R 3 is simply an ordered triple of real numbers. That is, 
a vector in R 3 may be written as 

(a\,a 2 , a 3 ) (e.g., (jt, e, <s/2)). 



To emphasize that we want to consider the pair or triple of numbers as a 
single unit, we will use boldface letters; hence a = (a\, a 2 ) or a = (ai, a%, a{) 
will be our standard notation for vectors in R 2 or R 3 . Whether we mean that a is a 
vector in R 2 or in R 3 will be clear from context (or else won't be important to the 
discussion). When doing handwritten work, it is difficult to "boldface" anything, 
so you'll want to put an arrow over the letter. Thus, a will mean the same thing 
as a. Whatever notation you decide to use, it's important that you distinguish the 
vector a (or a) from the single real number a. To contrast them with vectors, we 
will also refer to single real numbers as scalars. 

In order to do anything interesting with vectors, it's necessary to develop 
some arithmetic operations for working with them. Before doing this, however, 
we need to know when two vectors are equal. 



DEFINITION 1.2 Two vectors a = (ai,a 2 ) and b = (b i ,b 2 ) in R 2 are 
equal if their corresponding components are equal, that is, if a\ = b\ and 
a 2 = b 2 . The same definition holds for vectors in R 3 : a = (a\, a 2 , a^) and 
b = (b\, b 2 ,b-i) are equal if their corresponding components are equal, that 
is, if a\ =b\,a 2 = b 2 , and a^ = b^. 



2 Chapter 1 ] Vectors 

EXAMPLE 1 The vectors a = (1, 2) and b = (§, f ) are equal in R 2 , but c = 
(1, 2, 3) and d = (2, 3, 1) are not equal in R 3 . ♦ 

Next, we discuss the operations of vector addition and scalar multiplication. 
We'll do this by considering vectors in R 3 only; exactly the same remarks will 
hold for vectors in R 2 if we simply ignore the last component. 



DEFINITION 1.3 (Vector Addition) Let a = (a\, a 2 , 03) and b = (b\, 
b 2 ,b^) be two vectors in R 3 . Then the vector sum a + b is the vector in R 3 
obtained via componentwise addition: a 4- b = (a\ + b\, a 2 + b 2 , 03 + ^3). 



EXAMPLE 2 We have (0, 1,3) + (7, -2, 10) = (7, - 1, 13) and (in R 2 ): 

(1, 1) + (7T, V2) = (1 + 7T, 1 + V2). ♦ 

Properties of vector addition. We have 

1. a + b = b + a for all a, b in R 3 (commutativity); 

2. a + (b + c) = (a + b) + c for all a, b, c in R 3 (associativity); 

3. a special vector, denoted 0 (and called the zero vector), with the property 
that a + 0 = a for all a in R 3 . 



These three properties require proofs, which, like most facts involving the al- 
gebra of vectors, can be obtained by explicitly writing out the vector components. 
For example, for property 1, we have that if 

a = {a\,a 2 ,a{) and b = {b\, b 2 , ^3), 

then 

a + b = (czi + b\ , a 2 + b 2 , «3 + h) 
= (b\ + c/i , b 2 + a 2 , b-i + a 3 ) 
= b + a, 

since real number addition is commutative. For property 3, the "special vector" 
is just the vector whose components are all zero: 0 = (0, 0, 0). It's then easy to 
check that property 3 holds by writing out components. Similarly for property 2, 
so we leave the details as exercises. 



DEFINITION 1 .4 (Scalar Multiplication) Let a = (a\ , a 2 , 03) be a vec- 
tor in R 3 and let k e R be a scalar (real number). Then the scalar prod- 
uct ka is the vector in R 3 given by multiplying each component of a by 
k: ka = (ka\ , ka 2 , ka{). 



EXAMPLE 3 Ifa = (2,0, V2)and^ = 7,thenyta = (14, 0, 7V2). ♦ 

The results that follow are not difficult to check — just write out the vector 
components. 



1.1 | Vectors in Two and Three Dimensions 



Properties of scalar multiplication. For all vectors a and b in R 3 (or R 2 ) 

and scalars k and / in R, we have 

1. (k + /)a = ka + la (distributivity); 

2. k(a + b) = ka + kb (distributivity); 

3. k(la) = (kl)a = l(ka). 



It is worth remarking that none of these definitions or properties really de- 
pends on dimension, that is, on the number of components. Therefore we could 
have introduced the algebraic concept of a vector in R" as an ordered n-tuple 
(fli, d2, . . . , a n ) of real numbers and defined addition and scalar multiplication 
in a way analogous to what we did for R 2 and R 3 . Think about what such a 
generalization means. We will discuss some of the technicalities involved in § 1 .6. 

Vectors in R 2 and R 3 : The Geometric Notion 

Although the algebra of vectors is certainly important and you should become 
adept at working algebraically, the formal definitions and properties tend to present 
a rather sterile picture of vectors. A better motivation for the definitions just given 
comes from geometry. We explore this geometry now. First of all, the fact that 
a vector a in R 2 is a pair of real numbers (a\ , ai) should make you think of the 
coordinates of a point in R 2 . (See Figure 1.1.) Similarly, if a e R 3 , then a may 
be written as (a\, 02, as), and this triple of numbers may be thought of as the 
coordinates of a point in R 3 . (See Figure 1.2.) 

All of this is fine, but the results of performing vector addition or scalar mul- 
tiplication don't have very interesting or meaningful geometric interpretations in 
terms of points. As we shall see, it is better to visualize a vector in R 2 or R 3 as an 
arrow that begins at the origin and ends at the point. (See Figure 1.3.) Such a depic- 
tion is often referred to as the position vector of the point (a\ , ai) or {a\ , #2, ai). 

If you've studied vectors in physics, you have heard them described as objects 
having "magnitude and direction." Figure 1 .3 demonstrates this concept, provided 
that we take "magnitude" to mean "length of the arrow" and "direction" to be the 
orientation or sense of the arrow. (Note: There is an exception to this approach, 
namely, the zero vector. The zero vector just sits at the origin, like a point, and has 
no magnitude and therefore, an indeterminate direction. This exception will not 
pose much difficulty.) However, in physics, one doesn't demand that all vectors 




4 Chapter 1 | Vectors 



be represented by arrows having their tails bound to the origin. One is free to 
"parallel translate" vectors throughout R 2 and R 3 . That is, one may represent 
the vector a = (a\ , a 2 , a^) by an arrow with its tail at the origin (and its head at 
(fli, a 2 , 03)) or with its tail at any other point, so long as the length and sense of 
the arrow are not disturbed. (See Figure 1 .4.) For example, if we wish to represent 
a by an arrow with its tail at the point (jci, x 2 , ^3), then the head of the arrow 
would be at the point {x\ + ci\ , x 2 + a 2 , X3 + 03). (See Figure 1.5.) 




Figure 1 .4 Each arrow is a 
parallel translate of the position 
vector of the point (ai , 02, 03) and 
represents the same vector. 



(^i + flj, x 2 + a 2 , x 3 + a 3 ) 




(x h x 2 ,x 3 ) 



Figure 1 .5 The vector 

a = (a\ , ci2, 03) represented by an 

arrow with tail at the point 

(xi,x 2 , x 3 ). 




Figure 1 .6 The vector 
a + b may be represented 
by an arrow whose tail is at 
the tail of a and whose head 
is at the head of b. 



With this geometric description of vectors, vector addition can be visualized 
in two ways. The first is often referred to as the "head-to-tail" method for adding 
vectors. Draw the two vectors a and b to be added so that the tail of one of the 
vectors, say b, is at the head of the other. Then the vector sum a + b may be 
represented by an arrow whose tail is at the tail of a and whose head is at the head 
of b. (See Figure 1.6.) Note that it is not immediately obvious that a + b = b + a 
from this construction! 

The second way to visualize vector addition is according to the so-called 
parallelogram law: If a and b are nonparallel vectors drawn with their tails ema- 
nating from the same point, then a + b may be represented by the arrow (with its 
tail at the common initial point of a and b) that runs along a diagonal of the paral- 
lelogram determined by a and b (Figure 1.7). The parallelogram law is completely 
consistent with the head-to-tail method. To see why, just parallel translate b to the 
opposite side of the parallelogram. Then the diagonal just described is the result of 
adding a and (the translate of) b, using the head-to-tail method. (See Figure 1.8.) 

We still should check that these geometric constructions agree with our alge- 
braic definition. For simplicity, we'll work in R 2 . Let a = {a\ , a 2 ) and b = (b\ , b 2 ) 
as usual. Then the arrow obtained from the parallelogram law addition of a and 
b is the one whose tail is at the origin O and whose head is at the point P in 
Figure 1.9. If we parallel translate b so that its tail is at the head of a, then it is 
immediate that the coordinates of P must be (a\ + b\,a 2 + b 2 ), as desired. 

Scalar multiplication is easier to visualize: The vector ka. may be represented 
by an arrow whose length is \k\ times the length of a and whose direction is the 
same as that of a when k > 0 and the opposite when k < 0. (See Figure 1.10.) 

It is now a simple matter to obtain a geometric depiction of the difference 
between two vectors. (See Figure 1.11.) The difference a — b is nothing more 



1.1 | Vectors in Two and Three Dimensions 



a + b / 

/ 

/ 

i 



a + b 



(translated) 



Figure 1 .7 The vector 
a + b may be represented 
by the arrow that runs along 
the diagonal of the 
parallelogram determined 
by a and b. 



Figure 1 .8 The equivalence of the 
parallelogram law and the 
head-to-tail methods of vector 
addition. 




Figure 1.11 The 

geometry of vector 
subtraction. The vector c 
is such that b + c = a. 
Hence, c = a — b. 




2 < 



s_ — 




b / / b 














V v 






Figure 1 .9 The point P has coordinates 
(«i + b\, a 2 + b 2 ). 



Figure 1.10 Visualization of 
scalar multiplication. 



than a + (— b) (where — b means the scalar —1 times the vector b). The vector 
a — b may be represented by an arrow pointing from the head of b toward the 
head of a; such an arrow is also a diagonal of the parallelogram determined by a 
and b. ( As we have seen, the other diagonal can be used to represent a + b.) 
Here is a construction that will be useful to us from time to time. 



DEFINITION 1 .5 Given two points Pi(xi , yi , zi) and P 2 (x 2 , y 2 , z 2 ) in R 3 , 
the displacement vector from P\ to P 2 is 

P\A = (xi - x\, yi - yi, zi - zi). 



This construction is not hard to understand if we consider Figure 1.12. Given 
the points P\ and P 2 , draw the corresponding position vectors OP\ and OP 2 - 
Then we see that P\P 2 is precisely OP 2 —OP\. An analogous definition may 
~y be made for R 2 . 



/ In your study of the calculus of one variable, you no doubt used the notions of 

derivatives and integrals to look at such physical concepts as velocity, acceleration, 
Figure 1 .1 2 The displacement force, etc. The main drawback of the work you did was that the techniques involved 
vector P\P 2 , represented by the allowed you to study only rectilinear, or straight-line, activity. Intuitively, we all 
arrow from Pi to P 2 , is the understand that motion in the plane or in space is more complicated than straight- 
difference between the position line motion. Because vectors possess direction as well as magnitude, they are 
vectors of these two points. ideally suited for two- and three-dimensional dynamical problems. 



Chapter 1 | Vectors 




Figure 1.13 After t seconds, the 
point starting at a, with velocity v, 
moves to a + t\. 



Vj ship 
(with respect 
to still water) 



, current 



Net velocity 



Figure 1.14 The length of Vi is 
15, and the length of V2 is 5V2. 



For example, suppose a particle in space is at the point (ai, a.2, a{) (with 
respect to some appropriate coordinate system). Then it has position vector a = 
(a\, a 2 , 03). If the particle travels with constant velocity v = (uj, i>2, V3) for / 
seconds, then the particle's displacement from its original position is rv, and its 
new coordinate position is a + fv. (See Figure 1.13.) 

EXAMPLE 4 If a spaceship is at position (100, 3, 700) and is traveling with 
velocity (7, —10,25) (meaning that the ship travels 7 mi/sec in the positive 
x-direction, 10 mi/sec in the negative y-direction, and 25 mi/sec in the positive 
z-direction), then after 20 seconds, the ship will be at position 

(100, 3, 700) + 20(7, -10, 25) = (240, -197, 1200), 

and the displacement from the initial position is (140, —200, 500). ♦ 

EXAMPLE 5 The S.S. Calculus is cruising due south at a rate of 15 knots 
(nautical miles per hour) with respect to still water. However, there is also a 
current of 5^2 knots southeast. What is the total velocity of the ship? If the ship 
is initially at the origin and a lobster pot is at position (20, —79), will the ship 
collide with the lobster pot? 

Since velocities are vectors, the total velocity of the ship is Vi + V2, where vi is 
the velocity of the ship with respect to still water and V2 is the southeast-pointing 
velocity of the current. Figure 1.14 makes it fairly straightforward to compute 
these velocities. We have that vi = (0, —15). Since \2 points southeastward, its 
direction must be along the line v = — x. Therefore, V2 can be written as V2 = 
(v, —v), where v is a positive real number. By the Pythagorean theorem, if the 
length of \2 is 5\/2, then we must have v 2 + (— v) 2 = (5>/2) 2 or 2v 2 = 50, so 
that v = 5. Thus, v 2 = (5, —5), and hence, the net velocity is 

(0, -15) + (5, -5) = (5, -20). 
After 4 hours, therefore, the ship will be at position 

(0,0) + 4(5, -20) = (20, -80) 
and thus will miss the lobster pot. ♦ 

EXAMPLE 6 The theory behind the venerable martial art of judo is an excel- 
lent example of vector addition. If two people, one relatively strong and the other 
relatively weak, have a shoving match, it is clear who will prevail. For example, 
someone pushing one way with 200 lb of force will certainly succeed in overpow- 
ering another pushing the opposite way with 1 00 lb of force. Indeed, as Figure 1.15 
shows, the net force will be 100 lb in the direction in which the stronger person 
is pushing. 



1001b 



2001b 



1001b 



200 lb 




2001b 



Figure 1.16 Vector addition in 
judo. 



Figure 1 .1 5 A relatively strong person pushing with a 
force of 200 lb can quickly subdue a relatively weak one 
pushing with only 100 lb of force. 

Dr. Jigoro Kano, the founder of judo, realized (though he never expressed 
his idea in these terms) that this sort of vector addition favors the strong over the 
weak. However, if the weaker participant applies his or her 100 lb of force in a 
direction only slightly different from that of the stronger, he or she will effect a 
vector sum of length large enough to surprise the opponent. (See Figure 1.16.) 



1.1 | Exercises 



This is the basis for essentially all of the throws of judo and why judo is described 
as the art of "using a person's strength against himself or herself." In fact, the 
word "judo" means "the giving way." One "gives in" to the strength of another by 
attempting only to redirect his or her force rather than to oppose it. ♦ 



1.1 Exercises 



1 . Sketch the following vectors in R 2 : 

(a) (2,1) (b)(3,3) (c)(-l,2) 

2. Sketch the following vectors in R 3 : 

(a) (1,2,3) (b) (-2,0,2) (c) (2, -3, 1) 

3. Perform the indicated algebraic operations. Express 
your answers in the form of a single vector a = (a \ , 02) 
inR 2 . 

(a) (3,1) + (-1,7) 

(b) -2(8, 12) 

(c) (8,9) + 3(-l,2) 

(d) (1, 1) + 5(2, 6) -3(10, 2) 

(e) (8, 10) + 3 ((8, -2) - 2(4, 5)) 

4. Perform the indicated algebraic operations. Express 
your answers in the form of a single vector a = 
(ai , «2> a i) in R 3 - 

(a) (2, 1,2) + (-3, 9, 7) 

(b) 1(8,4, 1) + 2(5,-7, ±) 

(c) -2 ((2,0, l)-6(i,-4, 1)) 

5. Graph the vectors a = (1, 2), b = (-2, 5), and a + 
b = ( 1 , 2) + (—2, 5), using both the parallelogram law 
and the head-to-tail method. 

6. Graph the vectors a = (3, 2) and b = (— 1, 1). Also 
calculate and graph a — b, ia, and a + 2b. 

7. Let A be the point with coordinates (1,0, 2), let B be 
the point with coordinates (—3, 3, 1), and let C be the 
point with coordinates (2, 1,5). 

(a) Describe the vectors AB and BA. 

(b) Describe the vectors AC, B~C, and AC + CB. 

(c) Explain, with pictures, why AC + CB = AB. 

8. Graph (1,2, 1 ) and (0,-2, 3), and calculate and graph 
(1, 2, 1) + (0, -2, 3), -1(1,2, 1), and 4(1, 2, 1). 

9. If (-12, 9, z) + (x, 7, -3) = (2, y, 5), what are x, y, 
andz? 

10. What is the length (magnitude) of the vector (3, 1)? 
(Hint: A diagram will help.) 

11. Sketch the vectors a = (l,2)andb = (5, 10). Explain 
why a and b point in the same direction. 



12. Sketch the vectors a = (2, -7, 8) and b=(-l, 
j,— 4). Explain why a and b point in opposite 
directions. 

13. How would you add the vectors (1,2,3,4) and 
(5,-1, 2, 0) in R 4 ? What should 2(7, 6, -3, 1) be? In 
general, suppose that 

a = (a\ , <22, • ■ ■ , a„) and b = (bi, b%, ■ ■ ■ , b n ) 

are two vectors in R" and k e R is a scalar. Then how 
would you define a + b and fca? 

14. Find the displacement vectors from Pi to P2, where Pi 
and P2 are the points given. Sketch Pi, P2, and P1P2. 

(a) Pi(l,0,2),P 2 (2, 1,7) 

(b) Pi(l,6,-l),P 2 (0,4,2) 

(c) Pi(0,4,2),P 2 (l,6,-l) 

(d) Pi(3, 1),P 2 (2,-1) 

15. Let Pi(2, 5, -1, 6) and P 2 (3, 1, -2, 7) be two points 
in R 4 . How would you define and calculate the dis- 
placement vector from Pi to P2? (See Exercise 13.) 

16. If A is the point in R 3 with coordinates (2, 5, —6) and 
the displacement vector from A to a second point B is 
(12, —3, 7), what are the coordinates of Bl 

1 7. Suppose that you and your friend are in New York talk- 
ing on cellular phones. You inform each other of your 
own displacement vectors from the Empire State Build- 
ing to your current position. Explain how you can use 
this information to determine the displacement vector 
from you to your friend. 

1 8. Give the details of the proofs of properties 2 and 3 of 
vector addition given in this section. 

1 9. Prove the properties of scalar multiplication given in 
this section. 

20. (a) If a is a vector in R 2 or R 3 , what is 0a? Prove your 

answer. 

(b) If a is a vector in R 2 or R 3 , what is la? Prove your 
answer. 

21. (a) Let a = (2, 0) and b = (1, 1). For 0 < s < 1 and 

0 < t < 1, consider the vector x = sa + tb. Ex- 
plain why the vector x lies in the parallelogram 



8 Chapter 1 ] Vectors 



determined by a and b. (Hint: It may help to draw 
a picture.) 

(b) Now suppose that a = (2, 2, 1) and b = (0, 3, 2). 
Describe the set of vectors {x = sa + ?b | 0 < .y < 
1, 0 < t < 1}. 

22. Let a = {a\, ai, 03) andb = (b\, bi, bi) be two nonzero 
vectors such that b / ka. Use vectors to describe 
the set of points inside the parallelogram with vertex 
Po(xo, yo, zo) and whose adjacent sides are parallel to 
a and b and have the same lengths as a and b. (See 
Figure 1.17.) (Hint: If P(x, y, z) is a point in the par- 
allelogram, describe OP, the position vector of P.) 



z 




X 



Figure 1.17 Figure for Exercise 22. 



23. A flea falls onto marked graph paper at the point (3,2). 
She begins moving from that point with velocity vector 
v = (—1, —2) (i.e., she moves 1 graph paper unit per 
minute in the negative x -direction and 2 graph paper 
units per minute in the negative y-direction). 

(a) What is the speed of the flea? 

(b) Where is the flea after 3 minutes? 

(c) How long does it take the flea to get to the point 
(-4, -12)? 

(d) Does the flea reach the point (—13, —27)? Why or 
why not? 



24. A plane takes off from an airport with velocity vector 
(50, 100, 4). Assume that the units are miles per hour, 
that the positive a- -axis points east, and that the positive 
y-axis points north. 

(a) How fast is the plane climbing vertically at take- 
off? 

(b) Suppose the airport is located at the origin and a 
skyscraper is located 5 miles east and 10 miles 
north of the airport. The skyscraper is 1 ,250 feet tall. 
When will the plane be directly over the building? 

(c) When the plane is over the building, how much 
vertical clearance is there? 

25. As mentioned in the text, physical forces (e.g., gravity) 
are quantities possessing both magnitude and direction 
and therefore can be represented by vectors . If an obj ect 
has more than one force acting on it, then the resul- 
tant (or net) force can be represented by the sum of 
the individual force vectors. Suppose that two forces, 
F] = (2, 7, -1) and F 2 = (3, -2, 5), act on an object. 

(a) What is the resultant force of Fj and F2? 

(b) What force F3 is needed to counteract these forces 
(i.e., so that no net force results and the object 
remains at rest)? 

26. A 50 lb sandbag is suspended by two ropes. Suppose 
that a three-dimensional coordinate system is intro- 
duced so that the sandbag is at the origin and the ropes 
are anchored at the points (0, —2, 1) and (0, 2, 1). 

(a) Assuming that the force due to gravity points par- 
allel to the vector (0, 0,-1), give a vector F that 
describes this gravitational force. 

(b) Now, use vectors to describe the forces along each 
of the two ropes. Use symmetry considerations and 
draw a figure of the situation. 

27. A 10 lb weight is suspended in equilibrium by two 
ropes. Assume that the weight is at the point (1, 2, 3) 
in a three-dimensional coordinate system, where the 
positive z-axis points straight up, perpendicular to the 
ground, and that the ropes are anchored at the points 
(3,0,4) and (0,3,5). Give vectors Fj and F 2 that 
describe the forces along the ropes. 



1 .2 More About Vectors 

The Standard Basis Vectors 

In R 2 , the vectors i = (1, 0) and j = (0, 1) play a special notational role. Any 
vector a= (a\, Oq) may be written in terms of i and j via vector addition and 
scalar multiplication: 

a 2 ) — (fli, 0) + (0, a 2 ) = fli(l, 0) + a 2 (0, 1) = a x i + a 2 j. 

(It may be easier to follow this argument by reading it in reverse.) Insofar as nota- 
tion goes, the preceding work simply establishes that one can write either (a\ , a 2 ) 



1.2 | More About Vectors 



j 






i 



« 2 j 


----^^ a = a{\ + a 2 







Figure 1.18 Any vector in R 2 can be written in terms of i and j. 




Figure 1.19 Any vector in R can be written in terms of i, j, and k. 





v = 3 











Figure 1 .20 In R 2 , the equation 
y = 3 describes a line. 



or aji + ci2] to denote the vector a. It's your choice which notation to use (as long 
as you're consistent), but the ij-notation is generally useful for emphasizing the 
"vector" nature of a, while the coordinate notation is more useful for emphasizing 
the "point" nature of a (in the sense of a's role as a possible position vector of 
a point). Geometrically, the significance of the standard basis vectors i and j is 
that an arbitrary vector a e R 2 can be decomposed pictorially into appropriate 
vector components along the x- and y-axes, as shown in Figure 1.18. 

Exactly the same situation occurs in R 3 , except that we need three vec- 
tors, i = (1 , 0, 0), j = (0, 1 , 0), and k = (0, 0, 1), to form the standard basis. (See 
Figure 1.19.) The same argument as the one just given can be used to show that 
any vector a = (a\,a2, a^) may also be written as a\ i + aj. j + c?3 k. We shall 
use both coordinate and standard basis notation throughout this text. 




Figure 1 .21 In R 3 , the equation 
y = 3 describes a plane. 



EXAMPLE 1 We may write the vector (1, 
(7, n, —3) as 7i + 7ij — 3k. 



-2) as i — 2j and the vector 

♦ 



Parametric Equations of Lines 

In R 2 , we know that equations of the form y = mx + b or Ax + By = C describe 
straight lines. (See Figure 1.20.) Consequently, one might expect the same sort of 
equation to define a line in R 3 as well. Consideration of a simple example or two 
(such as in Figure 1.21) should convince you that a single such linear equation 
describes a plane, not a line. A pair of simultaneous equations in x, y, and z is 
required to define a line. 

We postpone discussing the derivation of equations for planes until §1.5 and 
concentrate here on using vectors to give sets of parametric equations for lines in 
R 2 orR 3 (or even R"). 



1 0 Chapter 1 | Vectors 




t = 3n:/2 



Figure 1 .22 The graph of the 
parametric equations x = 2 cos t, 
y = 2 sin t , 0 < t < 2n . 



First, we remark that a curve in the plane may be described analytically 
by points (x, y), where x and y are given as functions of a third variable (the 
parameter) t. These functions give rise to parametric equations for the curve: 

x = fit) 

y = g(t) ' 



EXAMPLE 2 The set of equations 

x = 2cosf 



0 < t < In 



y = 2 sin t 

describes a circle of radius 2, since we may check that 



x 2 + y 2 = (2 cos t) 2 + (2 sin t) 2 = 4. 



(See Figure 1.22.) 




Figure 1 .23 The line / is the 
unique line passing through Pq and 
parallel to the vector a. 




Parametric equations may be used as readily to describe curves in R 3 ; a curve 
in R 3 is the set of points (x, y, z) whose coordinates x,y, and z are each given by 
a function of t: 

x = f(t) 

y = g(t) ■ 

z. = hit) 

The advantages of using parametric equations are twofold. First, they offer a 
uniform way of describing curves in any number of dimensions. (How would 
you define parametric equations for a curve in R 4 ? In R 128 ?) Second, they allow 
you to get a dynamic sense of a curve if you consider the parameter variable t to 
represent time and imagine that a particle is traveling along the curve with time 
according to the given parametric equations. You can represent this geometrically 
by assigning a "direction" to the curve to signify increasing t. Notice the arrow 
in Figure 1.22. 

Now, we see how to provide equations for lines. First, convince yourself that 
a line in R 2 or R 3 is uniquely determined by two pieces of geometric information: 
(1) a vector whose direction is parallel to that of the line and (2) any particular 
point lying on the line — see Figure 1.23. In Figure 1.24, we seek the vector 

r = ~OP 

between the origin O and an arbitrary point P on the line / (i.e., the position 
vector of P(x, y, z))- OP is the vector sum of the position vector b of the given 
point P 0 (i.e., OPq) and a vector parallel to a. Any vector parallel to a must be a 
scalar multiple of a. Letting this scalar be the parameter variable t, we have 

r=0~P~ = OP^o + ta, 
and we have established the following proposition: 

PROPOSITION 2.1 The vector parametric equation for the line through the point 
Poib\ , b2, bi), whose position vector is OPq = b = b\i + b 2 ] + Z? 3 k, and parallel 
to a = aii + a2] + aj,k is 



Figure 1 .24 The graph of a line 
inR 3 . 



r(f) = b + /a. 



(1) 



1.2 | More About Vectors 11 



Expanding formula (1), 

r(f) = OP = b\i + b 2 j + & 3 k + t{a\\ + a 2 \ + o^k) 

= (ait + h)i + (a 2 t + b 2 )j + (a 3 t + 6 3 )k. 

Next, write op as xi + yj + zk so that P has coordinates (x, y, z). Then, ex- 
tracting components, we see that the coordinates of P are (a\t + b\, a 2 t + fe, 
+ b$) and our parametric equations are 

x = a\t + b\ 
■ y = a 2 t + b 2 , (2) 
z = a 3 t + Z? 3 

where f is any real number. 

These parametric equations work just as well in R 2 (if we ignore the z- 
component) or in R" where n is arbitrary. In R", formula (1) remains valid where 
we take a = (a,\, a 2 , . . . , a n ) and b = (b\, b 2 , . . . , b„). The resulting parametric 
equations are 

x\ = ci\t + b\ 
x 2 = a 2 t + b 2 

x n — ci n t -\- b n 




Figure 1.25 Finding equations 
for a line through two points in 
Example 4. 



EXAMPLE 3 To find the parametric equations of the line through (1, —2, 3)and 
parallel to the vector n\ — 3j + k, we have a = ni — 3j + k and b = i — 2j + 3k 
so that formula (1) yields 

r(/) = i - 2j + 3k + t(ni - 3j + k) 

= ( 1 + jtt)\ + (-2 - 3r)j + (3 + r)k 

The parametric equations may be read as 

X = 7tt + 1 

y = -3t-2 . 
v z = t + 3 ♦ 

EXAMPLE 4 From Euclidean geometry, two distinct points determine a unique 
line in R 2 or R 3 . Let's find the parametric equations of the line through the points 
Pq(\, —2, 3) and P\(0, 5, —1). The situation is suggested by Figure 1.25. To use 
formula (1), we need to find a vector a parallel to the desired line. The vector with 
tail at Pq and head at P\ is such a vector. That is, we may use for a the vector 



PoP^ = (0-1,5- (-2), -1 - 3) = -i + 7j - 4k. 

For b, the position vector of a particular point on the line, we have the choice 
of taking either b = i — 2j + 3k or b = 5j — k. Hence, the equations in (2) yield 
parametric equations 



x = 1 - t 
y = -2 + It 
z = 3 - At 



or 



x = — t 
y = 5 + It 

z = -1 -At 



Chapter 1 | Vectors 



In general, given two arbitrary points 

Po(a\, a 2 , a 3 ) and P\(b\, b 2 , fc 3 ), 
the line joining them has vector parametric equation 



r(t) = OP 0 + tP 0 Pi. 
Equation (3) gives parametric equations 

x = a\ + (b\ — a\)t 
y = a 2 + (b 2 - a 2 )t . 
Z = a 3 + (b2 - ai)t 

Alternatively, in place of equation (3), we could use the vector equation 



(3) 



(4) 



r(r) = OP\ + tP 0 P\ 



or perhaps 



r(r) = OP x + tP\P 0 



(5) 



(6) 



each of which gives rise to somewhat different sets of parametric equations. Again, 

we refer you to Figure 1 .25 for an understanding of the vector geometry involved. 

Example 4 brings up an important point, namely, that parametric equations 

for a line (or, more generally, for any curve) are never unique. In fact, the two 

sets of equations calculated in Example 4 are by no means the only ones; we 
> 

could have taken a = Pi Pq = i — 7j + 4k or any nonzero scalar multiple of 
P^p\ for a. 

If parametric equations are not determined uniquely, then how can you check 
your work? In general, this is not so easy to do, but in the case of lines, there are 
two approaches to take. One is to produce two points that lie on the line specified 
by the first set of parametric equations and see that these points lie on the line 
given by the second set of parametric equations. The other approach is to use the 
parametric equations to find what is called the symmetric form of a line in R 3 . 
From the equations in (2), assuming that each a, is nonzero, one can eliminate 
the parameter variable t in each equation to obtain: 

x — b\ 



t = 



t = 



y - b 2 
a 2 

z - h 
a 3 



The symmetric form is 



1.2 | More About Vectors 13 



In Example 4, the two sets of parametric equations give rise to corresponding 
symmetric forms 

x — 1 y + 2 z — 3 , x y — 5 z + 1 



■1 



and 



■1 



It's not difficult to see that adding 1 to each "side" of the second symmetric form 
yields the first one. In general, symmetric forms for lines can differ only by a 
constant term or constant scalar multiples (or both). 

The symmetric form is really a set of two simultaneous equations in R 3 . For 
example, the information in (7) can also be written as 

b\ y - b 2 



x 

fll 
x — b\ 



y 

fl3 



This illustrates that we require two "scalar" equations in x, y, and z to describe a 
line in R 3 , although a single vector parametric equation, formula (1), is sufficient. 

The next two examples illustrate how to use parametric equations for lines to 
identify the intersection of a line and a plane or of two lines. 

EXAMPLE 5 We find where the line with parametric equations 

x = t + 5 



y 



-2t - 4 
3t + 7 



intersects the plane 3x + 2y — lz = 2. 

To locate the point of intersection, we must find what value of the parameter t 
gives a point on the line that also lies in the plane. This is readily accomplished by 
substituting the parametric values for x, y, and z from the line into the equation 
for the plane 

3(r + 5) + 2(-2t - 4) - 7(3f + 7) = 2. (8) 

Solving equation (8) for t, we find that t = —2. Setting t equal to —2 in the 
parametric equations for the line yields the point (3,0, 1), which, indeed lies in 
the plane as well. ♦ 



EXAMPLE 6 We determine whether and where the two lines 

3t - 3 
and \ y = t 



x — t + 1 
y = 5t + 6 
z = -2t 



x 

y 

z = t+ 1 



intersect. 

The lines intersect provided that there is a specific value t\ for the parameter 
of the first line and a value h for the parameter of the second line that generate the 
same point. In other words, we must be able to find t\ and t 2 so that, by equating 
the respective parametric expressions for x, y, and z„ we have 



h + 1 = 3f 2 - 
5fi + 6 = t 2 
-2t { = t 2 + 1 



3 



(9) 



1 4 Chapter 1 | Vectors 



The last two equations of (9) yield 

t 2 = 5?i + 6 = -2*i - 1 => fi = — 1. 



Using t\ = — 1 in the second equation of (9), we find that t 2 = 1 . Note that the 
values t\ = — 1 and ?2 = 1 also satisfy the first equation of (9); therefore, we have 
solved the system. Setting t = — 1 in the set of parametric equations for the first 
line gives the desired intersection point, namely, (0, 1,2). ♦ 



Parametric Equations in General 

Vector geometry makes it relatively easy to find parametric equations for a variety 
of curves. We provide two examples. 

EXAMPLE 7 If a wheel rolls along a flat surface without slipping, a point on 
the rim of the wheel traces a curve called a cycloid, as shown in Figure 1.26. 




Figure 1 .27 The result of the 
wheel in Figure 1 .26 rolling 
through a central angle of t . 



D 








3x12 - A 








A 









Figure 1 .28 AP with its tail at 
the origin. 




Figure 1 .26 The graph of a cycloid. 

Suppose that the wheel has radius a and that coordinates in R 2 are chosen so that 
the point of interest on the wheel is initially at the origin. After the wheel has 
rolled through a central angle of t radians, the situation is as shown in Figure 1.27. 
We seek the vector OP, the position vector of P, in terms of the parameter t. 
Evidently, OP = OA + AP, where the point A is the center of the wheel. The 
vector O A is not difficult to determine. Its j-component must be a , since the center 
of the wheel does not vary vertically. Its i-component must equal the distance the 
wheel has rolled; if t is measured in radians, then this distance is at, the length 
of the arc of the circle having central angle t. Hence, OA = ati + aj. 

The value of vector methods becomes apparent when we determine A~P. 
Parallel translate the picture so that A~P has its tail at the origin, as in Figure 1 .28. 
From the parametric equations of a circle of radius a, 



aP = 



a cos 



3tt 

T 



+ a sin 



3tt 
T 



? I j = —a sin t\ — a cos t j, 



from the addition formulas for sine and cosine. We conclude that 

OP = 0~A + A~P = (ati + aj) + (—a sin ti — a cos t\) 
= a(t — sin t )i + a(l — cos t)j, 



1.2 | More About Vectors 15 



so the parametric equations are 

x = a(t — sinf) 

y = a(l — cos t) ♦ 

EXAMPLE 8 If you unwind adhesive tape from a nonrotating circular tape 
dispenser so that the unwound tape is held taut and tangent to the dispenser roll, 
then the end of the tape traces a curve called the involute of the circle. Let's 
find the parametric equations for this curve, assuming that the dispensing roll 
has constant radius a and is centered at the origin. (As more and more tape is 
unwound, the radius of the roll will, of course, decrease. We'll assume that little 
enough tape is unwound so that the radius of the roll remains constant.) 

Considering Figure 1 .29, we see that the position vector O P of the desired 
point P is the vector sum OB + BP. To determine O B and BP, we use the angle 
9 between the positive x-axis and OB as our parameter. Since B is a point on the 
circle, 

OB = a cos# i + a sin9 j. 





Unwound 
\ tape 




B / Involute 






\ 0 


J (a, 0) 



Figure 1 .29 Unwinding tape, as 
in Example 8. The point P 
describes a curve known as the 
involute of the circle. 




Figure 1 .30 The vector BP must 
make an angle of 9 — n/2 with the 
positive x-axis. 



To find the vector B P, parallel translate it so that its tail is at the origin. Figure 1 .30 
shows that B~P 's length must be a9, the amount of unwound tape, and its direction 
must be such that it makes an angle of 0 — n/2 with the positive x-axis. From our 
experience with circular geometry and, perhaps, polar coordinates, we see that 
BP is described by 



iO cos — |) i + ad sin (0 — — ^ j = a6 sin0 i — a9 cosO j. 

OP = ~0% + B~^ = a(cos6> +6>sin0)i + a(sin6» - 6>cos6>)j. 

Ix = a(cos9 + 0 sin#) 
y = a(sin 0 — 9 cos 0) 

are the parametric equations of the involute, whose graph is pictured in 
Figure 1.31. ♦ 



BP 
Hence, 

So 



Chapter 1 | Vectors 

1.2 Exercises 



In Exercises 1-5, write the given vector by using the standard 
basis vectors for R 2 and R 3 . 



1. (2,4) 2. (9,-6) 

4. (-1,2,5) 5. (2,4,0) 



3. (3,;r, -7) 



In Exercises 6—10, write the given vector without using the 
standard basis notation. 

6. i + j -3k 

7. 9i-2j + V2k 

8. -3(2i-7k) 

9. jri — j (Consider this to be a vector in R 2 .) 

10. jri — j (Consider this to be a vector in R 3 .) 

11. Letai =(1, 1) and a 2 = (1,-1). 

(a) Write the vector b = (3, 1) as ciai + C2a2, where 
c\ and C2 are appropriate scalars. 

(b) Repeat part (a) for the vector b = (3, — 5). 

(c) Show that any vector b = {b\, bj) in R 2 may be 
written in the form c^i + C2&2 for appropriate 
choices of the scalars c\, cj, (This shows that ai 
and a2 form a basis for R 2 that can be used instead 
of i and j.) 

12. Leta!=(l,0, -l),a 2 = (0, 1, 0), and a 3 =(1, 1,-1). 

(a) Find scalars c\, c%, C3, so as to write the vector 
b = (5, 6, —5) as Qai + C2a2 + c^. 

(b) Try to repeat part (a) for the vector b = (2, 3, 4). 
What happens? 

(c) Can the vectors ai, a2, a 3 be used as a basis for 
R 3 , instead of i, j, k? Why or why not? 

In Exercises 13—18, give a set of parametric equations for the 
lines so described. 

1 3. The line in R 3 through the point (2, — 1 , 5) that is par- 
allel to the vector i + 3j — 6k. 

14. The line in R 3 through the point (12, -2, 0) that is 
parallel to the vector 5i — 12j + k. 

1 5. The line in R 2 through the point (2, — 1 ) that is parallel 
to the vector i — 7j. 

16. The line in R 3 through the points (2,1,2) and 
(3,-1,5). 

17. The line in R 3 through the points (1,4,5) and 
(2,4,-1). 

18. The line in R 2 through the points (8, 5) and (1, 7). 

1 9. Write a set of parametric equations for the line in R 4 
through the point (1, 2, 0, 4) and parallel to the vector 
(-2,5,3,7). 



20. Write a set of parametric equations for the line in 
R 5 through the points (9,^,-1,5,2) and (-1,1, 
V2,7, 1). 

21 . (a) Write a set of parametric equations for the line in 

R 3 through the point (— 1 , 7, 3) and parallel to the 
vector 2i — j + 5k. 

(b) Write a set of parametric equations for the line 
through the points (5,-3, 4) and (0, 1, 9). 

(c) Write different (but equally correct) sets of equa- 
tions for parts (a) and (b). 

(d) Find the symmetric forms of your answers in 
(a)-(c). 

22. Give a symmetric form for the line having parametric 
equations x = 5 — 2t, y = 3t + 1, z = 6t — 4. 

23. Give a symmetric form for the line having parametric 
equations x = t + 1, y = 3t — 9, z = 6 — 8f . 

24. A certain line in R 3 has symmetric form 

x - 2 v-3 z+l 



5 -2 4 

Write a set of parametric equations for this line. 

25. Give a set of parametric equations for the line with 
symmetric form 

x + 5 y-1 z+10 



3 7-2 
26. Are the two lines with symmetric forms 
x - I y + 2 z+l 



and 



x - 4 y-1 z + 5 



10 -5 : 
the same? Why or why not? 
27. Show that the two sets of equations 



z x + 1 y + 6 z + 5 

- and = '- = 

5 _6 -14 -10 



x — 2 y — 1 

3 = 7 
actually represent the same line in R 3 . 

28. Determine whether the two lines / 1 and I2 defined by 
the sets of parametric equations h: x = 2t — 5, y = 
3t + 2, z = 1 - 6t, and l 2 : x = 1 - 2t, y = 11 - 3f, 
z = 6t — 17 are the same. (Hint: First find two points 
on l\ and then see if those points lie on I2.) 

29. Do the parametric equations l\\ x = 3t + 2, y = 
t - 7, z = 5t+l, and l 2 : x = 6t - 1, y = 2t - 8, 
z = 10? — 3 describe the same line? Why or why not? 



1.2 | Exercises 



30. Do the parametric equations x = 3f 3 + 7, y = 2 — f 3 , 
Z = 5/ 3 + 1 determine a line? Why or why not? 

31. Do the parametric equations x = 5t 2 — 1, y = 2t 2 + 
3, z = 1 — t 2 determine a line? Explain. 

32. A bird is flying along the straight-line path x = 2t + 7, 
y = t — 2, z= 1 — 3f, where/ is measured in minutes. 

(a) Where is the bird initially (at t = 0)? Where is the 
bird 3 minutes later? 

(b) Give a vector that is parallel to the bird's path. 

(c) When does the bird reach the point ( y , | , — y )? 

(d) Does the bird reach (17, 4, -14)? 

33. Find where the line x = 3t — 5, y = 2 — t, z = 6t in- 
tersects the plane x + 3y — z = 19. 

34. Where does the line x = 1 — At, y = t — 3/2, z = 
2t + 1 intersect the plane 5x — 2y + z = 1? 

35. Find the points of intersection of the line x = 2t — 3, 
y = 3t +2, z = 5 — t with each of the coordinate 
planes x = 0, y = 0, and z = 0. 

36. Show that the line x = 5 — t, y = 2t — 7, z = t — 3 is 
contained in the plane having equation 2x — y + 4z = 5. 

37. Does the line x = 5 — t, y = 2t — 3, z = It + 1 inter- 
sect the plane x — 3y + z = 1? Why? 

38. Find where the line having symmetric form 

x — 3_y + 2_z 
6 ~ 3 ~ 5 

intersects the plane with equation 2x — 5v + 3z + 8 = 0. 

39. Show that the line with symmetric form 

x -3 z+2 

^r = y - 5 = — 

lies entirely in the plane 3x + 3 y + z = 22. 

40. Does the line with symmetric form 

x+4 _ y-2 _ z-\ 
3 ~ -1 ~ -9 

intersect the plane 2x — 3y + z = 7? 

41 . Let a,b,cbe nonzero constants. Show that the line with 
parametric equations x = at + a, y = b, z = ct + c 
lies on the surface with equation x 2 /a 2 + y 2 /b 2 — 
Z 2 /c 2 = 1. 

42. Find the point of intersection of the two lines l\ : x = 
2t + 3, y = 3t + 3, z = 2t + 1 and l 2 :x = 15 - It, 
y = t-2,z = 3t-l. 

43. Do the lines l x :x = 2t+\, y = -3t, z = t-\ 
and l 2 :x = 3t + 1, y = t + 5, z = 7 -t intersect? 
Explain your answer. 



44. (a) Find the distance from the point (—2, 1, 5) to any 

point on the line x = 3t — 5, y = 1 — t,z = 4t + 7. 
(Your answer should be in terms of the param- 
eter t.) 

(b) Now find the distance between the point (—2, 1 , 5) 
and the line x = 3t - 5, y = 1 - t, z = At + 7. 
(The distance between a point and a line is the dis- 
tance between the given point and the closest point 
on the line.) 

45. (a) Describe the curve given parametrically by 

x = 2cos3r „ 27r 
0 < t < — . 
y = 2 sin3r 3 

What happens if we allow t to vary between 0 
and 27r? 

(b) Describe the curve given parametrically by 

x = 5cos3f „ 27r 
0 < t < — . 
y = 5 sin3r 3 

(c) Describe the curve given parametrically by 

{x = 5 sin3t „ 27r 
0 < t < — . 
y = 5 cos3f 3 

(d) Describe the curve given parametrically by 

x = 5 cos 3t „ 27r 
0 < t < — . 
y = 3 sin3r 3 

46. Suppose that a bicycle wheel of radius a rolls along a 
fiat surface without slipping. If a reflector is attached 
to a spoke of the wheel at a distance b from the center, 
the resulting curve traced by the reflector is called a 
curtate cycloid. One such cycloid appears in Fig- 
ure 1.32, where a = 3 and b = 2. 



y 





1 1 




1 

2ji 


i 

4?r 



Figure 1 .32 A curtate cycloid. 



Using vector methods or otherwise, find a set of para- 
metric equations for the curtate cycloid. Figure 1.33 
should help. (Take a low point of the cycloid to lie 



1 8 Chapter 1 | Vectors 




Figure 1 .34 A prolate cycloid. 




Figure 1 .33 The point P traces a 
curtate cycloid. 

on the v-axis.) There is no theoretical reason that the 
cycloid just described cannot have a < b, although in 
such case the bicycle-wheel-reflector application is no 
longer relevant. (When a < b, the parametrized curve 
that results is called a prolate cycloid.) Your paramet- 
ric equations should be such that the constants a and b 
can be chosen independently of one another. An exam- 
ple of a prolate cycloid, with a = 2 and b = 4, is shown 
in Figure 1.34. Try to think of a physical situation in 
which such a curve would arise. 



47. Egbert is unwinding tape from a circular dispenser of 
radius a by holding the tape taut and perpendicular to 
the dispenser. Find a set of parametric equations for 
the path traced by the end of the tape (the point P in 
Figure 1 .35) as Egbert unwinds the tape. Use the angle 
0 between OP and the positive x -axis for parameter. 
Assume that little enough tape is unwound so that the 
radius of the dispenser remains constant. 







p 

/ 




u 









Figure 1.35 Figure for Exercise 47. 



1.3 The Dot Product 

When we introduced the arithmetic notions of vector addition and scalar mul- 
tiplication, you may well have wondered why the product of two vectors was 
not defined. You might think that "vector multiplication" should be defined in a 
manner analogous to the way we defined vector addition (i.e., by componentwise 
multiplication). However, such a definition is not very useful. Instead we shall 
define and use two different concepts of a product of two vectors: (1) the Eu- 
clidean inner product, or "dot" product, which may be defined for two vectors in 
R" (where n is arbitrary) and (2) the "cross" or vector product, which is defined 
only for vectors in R 3 . 



1.3 | The Dot Product 19 
The Dot Product of Two Vectors 



DEFINITION 3.1 Let a = (a\,a 2 , a 3 ) and b = (b u b 2 , b 3 ) be two vectors 
in R 3 . The dot (or inner or scalar) product of a and b, denoted a • b, is 

a • b = a\b\ + a 2 b 2 + 03^3 ■ 

In R 2 , the analogous definition is 

a • b = a\b\ + a 2 b 2 , 

where a = (a\ , a 2 ) and b = (b\ , b 2 ). 



EXAMPLE 1 In R 3 , we have 

(1, -2, 5) • (2, 1, 3) = (1)(2) + (-2X1) + (5X3) = 15. 
(3i + 2j - k) • (i - 2k) = (3X1) + (2X0) + (-1X-2) = 5. # 

In accordance with its name, the dot — or scalar — product takes two vectors 
and produces a single real number (not a vector). 

The following facts are consequences of Definition 3.1: 



Properties of dot products. If a, b, and c are any vectors in R 3 (or R 2 ) and 
k e R is any scalar, then 

1. a • a > 0, and a • a = 0 if and only if a = 0; 

2. a-b = ba; 

3. a • (b + c) = a • b + a • c; 

4. (jfca) • b = jfc(a • b) = a • (kb). 



Proof of Property 1 If a = (a\, a 2 , a^), then we have 

a • a = a\ci\ + a 2 a 2 + aj,a^ = a\ + a\+ a\. 

This last expression is evidently nonnegative, since it is a sum of squares of real 
numbers. Moreover, such an expression is zero exactly when each of the terms is 
zero, that is, if and only if a\ = a 2 = = 0. ■ 

We leave the proofs of properties 2, 3, and 4 as exercises. 

Thus far, we have introduced the dot product of two vectors as a purely 
algebraic construction. It is the geometric interpretation of the definition that is 
really interesting. To establish this interpretation, we begin with the following: 



DEFINITION 3.2 If a = (a\,a 2 , a 3 ), then the length of a (also called the 
norm or magnitude), denoted ||a||, is ^/a 2 x + a 2 + a 2 . 



Chapter 1 | Vectors 




a 



Figure 1.36 The dot 

product of a and b is 
llall llbll cosd. 




a 



Figure 1 .37 The vector 
triangle used in the proof 
of Theorem 3.3. 



The motivation for this definition is evident if we draw a as the position vector of 
the point (a\ , a%, a$). Then the length of the arrow from the origin to {a\ , 02, 03) is 

V(ai - 0) 2 + (a 2 " 0) 2 + (a 3 - 0) 2 , 

as given by the distance formula, which is nothing more than an extension of the 
Pythagorean theorem in the plane. As we just saw, a ■ a = a\ + a\ + a\, and we 
have 

||a|| = v / a T a 

or, equivalently, 



a - a = ||a|| 2 . (1) 



Now we're ready to state the main result concerning the geometry of the dot 
product. If a and b are two nonzero vectors in R 3 (or R 2 ) drawn with their tails at 
the same point, let 9, where 0 < 9 < it, be the angle between a and b. If either a 
or b is the zero vector, then 9 is indeterminate (i.e., can be any angle). 

THEOREM 3.3 If a and b are any two vectors in either R 2 or R 3 , then 

a-b= ||a|| ||b|| cos (9. 

(See Figure 1.36.) 



PROOF If either a or b is the zero vector, say a, then a = (0, 0, 0) and so 

a.b = (0)(&i) + (0)(&2) + (0)0>3) = 0. 
Also, || a|| = 0 in this case, so the formula in Theorem 3.3 holds. In this case, the 
angle 6 is indeterminate. 

Now suppose that neither a nor b is the zero vector. Let c = b — a. Then 
we may apply the law of cosines to the triangle whose sides are a, b, and c 
(Figure 1.37) to obtain 

Hell 2 = ||a|| 2 + ||b|| 2 -2||a|| ||b|| cose. 

Thus, 

2 1| a|| ||b|| cose = ||a|| 2 + ||b|| 2 - ||c|| 2 = a • a + b • b - c • c, (2) 
from equation (1). Now, use the properties of the dot product. Since c = b — a, 
c • c = (b - a) • (b — a) 

= (b - a) • b - (b - a) • a 

= b- b- a- b- b- a + a-a, (3) 

by properties 3 and 4 of the dot product. If we use equation (3) to substitute for 
c • c in equation (2), then 

2 1| a || ||b|| cos6» = a-a + b-b-(b-b-a-b-b-a + a-a) 

= a • b + b • a 

= 2a-b, 

by property 2 of the dot product. By canceling the factor of 2 on both sides, the 
desired result is obtained. ■ 



1.3 | The Dot Product 21 




Figure 1 .38 An object 
sliding down a ramp. 
The force due to gravity 
is downward, but the 
direction of travel of the 
object is inclined 30° to 
the horizontal. 



Angles Between Vectors 

Theorem 3.3 may be used to find the angle between two nonzero vectors a and 
b — just solve for 6 in the formula in Theorem 3.3 to obtain 




The use of the inverse cosine is unambiguous, since we take 0 < 6 < it when 
defining angles between vectors. 



EXAMPLE 2 If a = i + j and b = j - k, then formula (4) gives 



= cos 



-i (i + j)-(j 



= cos 



1 



= COS 



-1 



1 



7T 

3' 



+ jllllj-k|| (V2-V2) 

If a and b are nonzero, then Theorem 3.3 implies 

cos 6 = 0 if and only if a • b = 0. 

We have cos 6> = Ojust incased = jt/2. (Remember our restriction on 0.) Hence, 
it makes sense for us to call a and b perpendicular (or orthogonal) when a • b = 
0. If either a or b is the zero vector, then we cannot use formula (4), and the angle 
9 is undefined. Nonetheless, since a • b = 0 if a or b is 0, we adopt the standard 
convention and say that the zero vector is perpendicular to every vector. 

EXAMPLE 3 The vector i + j is orthogonal to the vector i — j + k, since 

(i + j) • (i - j + k) = (1X1) + (IX- 1) + (0X1) = 0- ♦ 



Vector Projections 

Suppose that a 2 kg object is sliding down a ramp having a 30° incline with the 
horizontal as in Figure 1.38. If we neglect friction, the only force acting on the 
object is gravity. What is the component of the gravitational force in the direction 
of motion of the object? 

To answer questions of this nature, we need to find the projection of one vector 
on another. The general idea is as follows: Given two nonzero vectors a and b, 
imagine dropping a perpendicular line from the head of b to the line through a. 
Then the projection of b onto a, denoted proj a b, is the vector represented by the 
arrow in Figure 1.39. 





proj a b proj a b 
Figure 1 .39 Projection of the vector b onto the vector a. 



22 Chapter 1 | Vectors 



Given this intuitive understanding of the projection, we find a precise formula 
for it. Recall that a vector is determined by magnitude (length) and direction. It 
follows by definition that the direction of proj a b is either the same as that of a, 
or opposite to a if the angle 9 between a and b is more than jt/2. Trigonometry 
then tells us 

, m II P ro jab || 
cos# = - -. 



(The absolute value sign around cos# is needed in case jt/2 < 0 < jt.) Hence, 
with a bit of algebra, we have 

II -kII mi .I I m ||a|| ||b|||cos0| |a-b| 
||pro Ja b|| = ||b|||cos 0 | = = — 

by Theorem 3.3. Thus, we know the magnitude and direction of proj a b. To obtain 
a compact formula for proj a b, note the following: 

PROPOSITION 3.4 Let k be any scalar and a any vector. Then 

1. II ka\\ = \k\ ||a||. 

2. A unit vector (i.e., a vector of length 1) in the direction of a nonzero vector 
a is given by a/||a||. 



PROOF Part 1 is left as an exercise. (Write out ka and ||&a|| in terms of compo- 
nents.) For part 2, we must check that the length of a/||a|| is 1: 



a 




1 






a 


Ilall 




Ilall 



|a|| = 1, 



by part 1 (since l/||a|| is a positive scalar). 

Now proj a b is a vector of length a • b|, 

|a-b|\ a 
Tal 



projab = ± 



| a || in the "±a-direction." That is, 
_ ± ||a|| ||b|| |cosfj| a 



length of 
proj a b 



unit vector 
in direction of a 



Note that the angle 9 keeps track of the appropriate sign of proj a b; that is, when 
0 < 6 < jt/2, cos 0 is positive and proj a b points in the direction of a, and when 
7r/2 < 8 < Tt, cos 9 is negative and proj a b points in the direction opposite to that 
of a. Thus, we can eliminate both the ± sign and the absolute value, and we find 
that 

||a|| ||b|| cos<9 a a-b 

proi„b = = r a 

Ilall ||a|| ||a|| 2 

by Theorem 3.3, so that 




by equation (1). Formula (5) is concise and not difficult to remember. 



1.3 | The Dot Product 




Figure 1 .40 The 2 kg 

object sliding down a 
ramp in Example 4. 



EXAMPLE 4 To answer the question posed at the beginning of this subsection, 
we need to calculate proj a F, where F is the gravitational force vector and a points 
along the ramp as shown in Figure 1 .40. We have a coordinate situation as shown 
in Figure 1.41. From trigonometric considerations, we must have a = a\\ + a^j 
such that flj = — ||a|| cos 30° and a 2 = — ||a|| sin 30°. Since we are really only 
interested in the direction of a, there is no loss in assuming that a is a unit vector. 
Thus, 

a = - cos 30° i- sin 30° j = -^i- \\. 







al^b° 




F = -mgj = -19.6j 





Figure 1 .41 The vectors a and F in Example 4, 
realized in a coordinate system. 



Taking g = 9.8 m/sec , we have F 
implies 



-2gj = — 19. 6j. Therefore, formula (5) 



. a-F 

proj a F = | | a = 



a • a 



(- |j)-(-19.6j) / V3. O 




27 

« -8.49i - 4.9 j, 
and the component of F in this direction is 

||proj a F|| = ||-8.49i-4.9j|| =9.8N. ♦ 

Unit vectors — that is, vectors of length 1 — are important in that they capture 
the idea of direction (since they all have the same length). Part 2 of Proposition 
3.4 shows that every nonzero vector a can have its length adjusted to give a unit 
vector u = a/ 1| a|| that points in the same direction as a. This operation is referred 
to as normalization of the vector a. 

EXAMPLE 5 A fluid is flowing across a plane surface with uniform velocity 
vector v. If n is a unit vector perpendicular to the plane surface, let's find (in terms 
of v and n) the volume of the fluid that passes through a unit area of the plane in 
unit time. (See Figure 1.42.) 



24 Chapter 1 | Vectors 




Figure 1 .42 Fluid flowing across 
a plane surface. 




Base has area 1 

Figure 1 .43 After one unit of time, the fluid passing 
across a square will have filled the box. 



First, imagine one unit of time has elapsed. Then over a unit area of the plane 
(say over a unit square), the fluid will have filled a "box" as in Figure 1.43. The 
box may be represented by a parallelepiped (a three-dimensional analogue of a 
parallelogram). The volume we seek is the volume of this parallelepiped and is 

Volume = (area of base)(height). 

The area of the base is 1 unit by construction. The height is given by ||proj n v||. 
From formula (5), 

/n-v\ 

P r °Jn v = ( ) n = (n • v)n, 

vn • n/ 

since n • n = ||n|| 2 = 1. Hence, 

||proj n v|| = ||(n-v)n|| = |n-v| ||n|| = |n-v|, 
by part 1 of Proposition 3.4. ♦ 



Vector Proofs 

We conclude this section with two illustrations of how wonderfully well vectors 
are suited to providing elegant proofs of geometric results. 

EXAMPLE 6 In an arbitrary triangle, show that the line segment joining the 
midpoints of two sides is parallel to and has half the length of the third side. (See 
Figure 1.44.) In other words, if M\ is the midpoint of side AB and M 2 is the 
midpoint of side AC, we wish to show that M\ M 2 is parallel to BC and has half 
its length. 



A A 




Figure 1 .44 In triangle ABC, Figure 1 .45 The vector version 

Mi M 2 is parallel to B C and has of triangle ABC in Example 6. 

half its length. 

For a vector proof, we use the diagram in Figure 1 .45, a slightly modified ver- 
sion of Figure 1 .44. The midpoint conditions translate to the following statements 
about vectors: 

AA?i = ±A$, A~\f 2 = \At. 



1.3 | The Dot Product 



Now, 

M1M2 = AM 2 - ~AM\ = \AC - \lh = \{AC - AB) = \~B~£. 

But M\ M2 = \BC is precisely what we wish to prove: To say M\M2 is a scalar 

times BC means that the two vectors are parallel. Moreover, from part 1 of 
Proposition 3.4, 

\\Mjt 2 \\ = \\\BC\\ = I||fi£||, 
so that the length condition also holds. ♦ 

EXAMPLE 7 Show that every angle inscribed in a semicircle is a right angle, 
as suggested by Figure 1 .46. 




Figure 1.46 Every angle Figure 1.47 a and b are "radius 

inscribed in a semicircle is a vectors." 
right angle. 



To prove this remark, we'll make use of Figure 1 .47, where a and b are "radius 
vectors" with tails at the center of the circle. We need only show that a — b (a 
vector along one ray of the angle in question) is perpendicular to — a — b (a vector 
along the other ray). In other words, we wish to show that 

(a - b) • (-a - b) = 0. 

We have 

(a - b) • (-a - b) = (-l)(a - b) • (a + b), 
by property 4 of dot products, 

= (-l)((a-b)-a + (a-b)-b) 
= (-l)(a a-b a + a b-b b) 
= (-l)(||a|| 2 - ||b|| 2 ), 

by properties 2 and 4, 

= 0, 

since both a and b are radius vectors (and therefore have the same length, namely, 
the radius of the circle). ♦ 

Vector proofs as in Examples 6 and 7 are elegant and sometimes allow you to 
write shorter and more direct proofs than those from your high school geometry 
days. 



26 Chapter 1 [ Vectors 

1.3 Exercises 



Compute a • b, ||a||, ||b|| for the vectors listed in Exercises 1—6. 



1. 


a = 


(l,5),b = (-2,3) 


2. 


a = 


(4,-l),b = (±,2) 


3. 


a = 


(-1,0, 7), b = (2, 4, -6) 


4. 


a = 


(2, 1, 0), b = (1, -2, 3) 


5. 


a = 


4i - 3j + k, b = i + j + k 


6. 


a = 


i + 2j-k,b = -3j + 2k 



In Exercises 7-11, find the angle between each of the pairs of 
vectors. 



7. 


a = 


V3i + j,b = — a/3 i + j 


8. 


a 


(-1,2), b = (3, 1) 


9. 


a = 


i + j,b = i + j + k 


10. 


a = 


i + j - k, b = -i + 2j + 2k 


11. 


a = 


(l,-2,3),b = (3,-6, -5) 



In Exercises 12—16, calculate proj a b. 

12. a = i + j, b = 2i + 3j - k 

13. a = (i + j)/V2, b = 2i + 3j-k 

14. a = 5k, b = i-j + 2k 

15. a= -3k,b = i-j + 2k 

16. a = i + j + 2k, b = 2i - 4j + k 

1 7. Give a unit vector that points in the same direction as 
the vector 2i — j + k. 

1 8. Give a unit vector that points in the direction opposite 
to the vector — i + 2k. 

1 9. Give a vector of length 3 that points in the same direc- 
tion as the vector i + j — k. 

20. Find three nonparallel vectors that are perpendicular 
to i - j + k. 

21. Is it ever the case that proj a b = proj b a? If so, under 
what conditions? 

22. Prove properties 2, 3, and 4 of dot products. 

23. Prove part 1 of Proposition 3.4. 

24. Suppose that a force F = i — 2j is acting on an object 
moving parallel to the vector a = 4i + j. Decompose F 
into a sum of vectors Fi and F2, where F ; points along 
the direction of motion and F2 is perpendicular to the 
direction of motion. (Hint: A diagram may help.) 



25. In physics, when a constant force acts on an object 
as the object is displaced, the work done by the force 
is the product of the length of the displacement and 
the component of the force in the direction of the dis- 
placement. Figure 1 .48 depicts an object acted upon by 
a constant force F, which displaces it from the point P 
to the point Q. Let 6 denote the angle between F and 
the direction of displacement. 

(a) Show that the work done by F is determined by the 
formula F- Pg. 

(b) Find the work done by the (constant) force F = 
i + 5j + 2k in moving a particle from the point 
(1, -1, 1) to the point (2, 0, -1). 




F 

6 L 



>Q 



Component of F in direction 
of displacement 

Figure 1.48 A constant force F 
displaces the object from P to Q. (See 
Exercise 25.) 

26. A refrigerator is dragged 12 ft across a smooth floor 
using a rope and 60 lb of force directed along the rope. 
How much work is done if the rope makes a 20° angle 
with the horizontal? 

27. How much work is done in pushing a handtruck loaded 
with 500 lb of bananas 40 ft up a ramp inclined 30° 
from horizontal? 

Let a be a nonzero vector in R 3 . The direction cosines of a are 

the three numbers cos a, cos ft, cos y determined by the angles 
a, fi, y between a and, respectively, the positive x-, y-, and 
Z-axes. In Exercises 28 and 29, find the direction cosines of the 
given vectors. 



28. a 

29. a 



i + 2j-k 
3i + 4k 



30. If a = flii + 02) + 03k, give expressions for the direc- 
tion cosines of a in terms of the components of a. 

31. Let A, P, and C denote the vertices of a triangle. Let 
0 < r < 1 . If Pi is the point on AS located r times the 
distance from A to 5 and Pi is the point on AC located 
r times the distance from A to C, use vectors to show 
that Pi P2 is parallel to PC and has r times the length 
of PC. (This result generalizes that of Example 6 of 
this section.) 



1.4 | The Cross Product 



32. Let A, B, C, and D be four points in R 3 such that no 
three of them lie on a line. Then ABCD is a quadri- 
lateral, though not necessarily one that lies in a plane. 
Denote the midpoints of the four sides of ABCD by 
Mi, M%, M3, and M4. Use vectors to show that, amaz- 
ingly, M1M2M2M4 is always a parallelogram. 

33. Use vectors to show that the diagonals of a parallel- 
ogram have the same length if and only if the paral- 
lelogram is a rectangle. (Hint: Let a and b be vectors 
along two sides of the parallelogram. Express vectors 
running along the diagonals in terms of a and b. See 
Figure 1.49.) 




a 



Figure 1.49 Diagram for Exercise 33. 

34. Using vectors, prove that the diagonals of a parallelo- 
gram are perpendicular if and only if the parallelogram 
is a rhombus. (Note: A rhombus is a parallelogram 
whose four sides all have the same length.) 

35. This problem concerns three circles of equal radius r 
that intersect in a single point O. (See Figure 1.50.) 

(a) Let W\, W2, and W3 denote the centers of the 

three circles and let w,- = OWi for i = 1, 2, 3. 
Similarly, let A, B, and C denote the remaining 

intersection points of the circles and set a = OA, 
b = OB, and c = OC. By numbering the centers 
of the circles appropriately, write a, b, and c in 
terms of wi , W2, and W3 . 

(b) Show that A, B, and C lie on a circle of the 
same radius r as the three given circles. (Hint: 



The center of the circle is at the point P, where 

OP = Wl + W2 + W3 .) 




Figure 1.50 Two examples 
of three circles of equal radius 
intersecting in a single 
point O. (See Exercise 35.) 



(c) Show that O is the orthocenter of triangle ABC. 
(The orthocenter of a triangle is the common in- 
tersection point of the altitudes perpendicular to 
the edges.) 

36. (a) Show that the vectors [|b[|a+ [|a||b and ||b||a — 
|| a || b are orthogonal. 

(b) Show that ||b||a + ||a||b bisects the angle between 
a and b. 



1 .4 The Cross Product 

The cross product of two vectors in R 3 is an "honest" product in the sense that it 
takes two vectors and produces a third one. However, the cross product possesses 
some curious properties (not the least of which is that it cannot be defined for 
vectors in R 2 without first embedding them in R 3 in some way) making it less 
"natural" than may at first seem to be the case. 

When we defined the concepts of vector addition, scalar multiplication, and 
the dot product, we did so algebraically (i.e., by a formula in the vector compo- 
nents) and then saw what these definitions meant geometrically. In contrast, we 
will define the cross product first geometrically, and then deduce an algebraic for- 
mula for it. This technique is more convenient, since the coordinate formulation 



28 Chapter 1 | Vectors 




a 

Figure 1 .51 The area of this 
parallelogram is ||a|| ||b|| sin#. 




Figure 1 .53 i x j = k. 




Figure 1 .54 A mnemonic for 
finding the cross product of the 
unit basis vectors. 



is fairly complicated (although we will find a way to organize it so as to make it 
easier to remember). 



The Cross Product of Two Vectors in R 3 



DEFINITION 4.1 Let a and b be two vectors in R 3 (not R 2 ). The cross 
product (or vector product) of a and b, denoted a x b, is the vector whose 
length and direction are given as follows: 

• The length of a x b is the area of the parallelogram spanned by a and b 
or is zero if either a is parallel to b or if a or b is 0. Alternatively, the 
following formula holds: 

||axb|| = ||a|| ||b|| sin 6, 

where 0 is the angle between a and b. (See Figure 1.51.) 

• The direction of a x b is such that a x b is perpendicular to both a and 
b (when both a and b are nonzero) and is taken so that the ordered triple 
(a, b, a x b) is a right-handed set of vectors, as shown in Figure 1.52. 
(If either a or b is 0, or if a is parallel to b, then a x b = 0 from the 
aforementioned length condition.) 

By saying that (a, b, a x b) is right-handed we mean that if you let the fingers 
of your right hand curl from a toward b, then your thumb will point in the 
direction of a x b. 



EXAMPLE 1 Let's compute the cross product of the standard basis vectors 
for R 3 . First consider i x j as shown in Figure 1.53. The vectors i and j deter- 
mine a square of unit area. Thus, ||i x j|| = 1. Any vector perpendicular to both 
i and j must be perpendicular to the plane in which i and j lie. Hence, i x j 
must point in the direction of ±k. The "right-hand rule" implies that i x j must 
point in the positive k direction. Since ||k|| = 1, we conclude that i x j = k. The 
same argument establishes that j x k = i and k x i = j. To remember these ba- 
sic equations, you can draw i, j, and k in a circle, as in Figure 1.54. Then the 
relations 



ixj = k, jxk = i, kxi = j (1) 



may be read from the circle by beginning at any vector and then proceeding 
clockwise. ♦ 



Properties of the Cross Product; Coordinate Formula 

Example 1 demonstrates that the calculation of cross products from the geometric 
definition is not entirely routine. What we really need is a coordinate formula, 
analogous to that for the dot product or for vector projections, which is not difficult 
to obtain. 



1.4 I The Cross Product 29 
From our Definition 4.1, it is possible to establish the following: 



Properties of the Cross Product. Let a, b, and c be any three vectors in R 3 
and let k e R be any scalar. Then 

1. axb = — bxa (anticommutativity); 

2. ax(b + c) = axb + axc (distributivity); 

3. (a + b)xc = axc + bxc (distributivity); 

4. k(a x b) = (fca) x b = a x (kb). 



We provide proofs of these properties at the end of the section, although you 
might give some thought now as to why they hold. It's worth remarking that these 
properties are entirely reasonable, ones that we would certainly want a product to 
have. However, you should be clear about the fact that the cross product fails to 
satisfy other properties that you might also consider to be eminently reasonable. 
In particular, since property 1 holds, we see that axb/bxain general (i.e., 
the cross product is not commutative). Consequently, be very careful about the 
order in which you write cross products. Another property that the cross product 
does not possess is associativity. That is, 

a x (b x c) / (a x b) x c, 

in general. For example, let a = b = i and c = j. Then 

i x (i x j) = i x k = -k x i = -j, 

from properties 1 and 4, but (ixi)xj = 0xj = 0^— j. (The equation 
i x i = 0 holds because i is, of course, parallel to i.) Make sure that you do 
not try to use an associative law when working problems. 

We now have the tools for producing a coordinate formula for the cross 
product. Let a = a\i + a 2 \ + a^k and b = b\\ + b 2 \ + b^k. Then 

a x b = (a\i + a 2 \ + a^k) x (b\\ + b 2 \ + b^k) 

= (aii + a 2 \ + 03k) x bii + (flii + a 2 \ + a 3 k) x b 2 j 

+ (aii + a 2 ) + 03k) x Z? 3 k, 

by property 2, 

= aib\\ x i + a 2 b\j x i + c^ik x i + a\b 2 \ x j + a 2 b 2 \ x j 
+ a^b 2 k x j + aib^i x k + a 2 b^i x k + a^b^k x k, 

by properties 3 and 4. These nine terms may look rather formidable at first, but 
we can simplify by means of the formulas in (1), anticommutativity, and the fact 
that c x c = 0 for any vector c e R 3 . (Why?) Thus, 

a x b = —a 2 b\k + «3i>ij + a\b 2 k — a^b 2 i — a\bi] + #2^31 

= (a 2 b 3 - a 3 b 2 )\ + (a 3 bi - + {a x b 2 - a 2 b x )k. (2) 

EXAMPLE 2 Formula (2) gives 

(i + 3j - 2k) x (2i + 2k) = (3 • 2 - (-2) • 0)i + (-2 • 2 - 1 • 2)j 

+ (1 • 0 - 3 • 2)k 
= 6i - 6j - 6k. ♦ 



Chapter 1 | Vectors 



Formula (2) is more complicated than the corresponding formulas for all the 
other arithmetic operations of vectors that we've seen. Moreover, it is a rather dif- 
ficult formula to remember. Fortunately, there is a more elegant way to understand 
formula (2). We explore this reformulation next. 



Matrices and Determinants: A First Introduction — 
A matrix is a rectangular array of numbers. Examples of matrices are 



1 2 3 
4 5 6 



"l 


3 




2 


7 


and 


0 


0 





1 0 0 0" 

0 10 0 

0 0 10 

0 0 0 1 



If a matrix has m rows and n columns, we call it "m x n" (read "m by w"). 
Thus, the three matrices just mentioned are, respectively, 2 x 3,3 x 2, and 4 x 4. 
To some extent, matrices behave algebraically like vectors. We discuss some 
elementary matrix algebra in § 1 .6. For now, we are mainly interested in the notion 
of a determinant, which is a real number associated toann x « (square) matrix. 
(There is no such thing as the determinant of a nonsquare matrix.) In fact, for the 
purposes of understanding the cross product, we need only study 2x2 and 3x3 
determinants. 



DEFINITION 4.2 Let A be a 2 x 2 or 3 x 3 matrix. Then the determinant 

of A, denoted det A or | A|, is the real number computed from the individual 
entries of A as follows: 



• 2 x 2 case 
If A = 

• 3 x 3 case 



, then |A| = 





a 


b c 








d 


e f 


, then 




J 


h i 










a b 


c 




\A 




d e 


f 








g h 


i 





= ad — be. 



= aei +bfg + cdh — ceg — afh — bdi 



e f 
h i 



d f 
g i 



+ c 



d e 
g h 



in terms of 2 x 2 determinants. 



Perhaps the easiest way to remember and compute 2x2 and 3x3 determi- 
nants (but not higher-order determinants) is by means of a "diagonal approach." 
We write (or imagine) diagonal lines running through the matrix entries. The 
determinant is the sum of the products of the entries that lie on the same diagonal, 



1.4 | The Cross Product 



where negative signs are inserted in front of the products arising from diagonals 
going from lower left to upper right: 



•2x2 case 



A = 




and 



\A\ = ad — be. 

3x3 case (we need to repeat the first two columns for the method to work) 

a b c 

d e f 

_ g h i 

Write 



A = 




Then 



\A\ = aei + bfg + cdh — gee — hfa — idb. 



Important Warning This mnemonic device does not generalize beyond 
3x3 determinants. 

We now state the connection between determinants and cross products. 



Key Fact. If a = a\\ + d2] + a^k and b = b\i + bi\ + ^k, then 

j + 



a x b 



a 2 a 3 
b 2 h 



bx b 3 



b\ b 2 



k = 



i j k 

a,\ a 2 «3 
b\ b 2 b} 



(3) 



The determinants arise from nothing more than rewriting formula (2). Note 
that the 3 x 3 determinant in formula (3) needs to be interpreted by using the 2 x 2 
determinants that appear in formula (3). (The 3x3 determinant is sometimes 
referred to as a "symbolic determinant.") 



EXAMPLE 3 

(3i + 2j-k)x(i-j + k) 



2 -1 




3 -1 


j + 


3 2 


1 1 


i — 


1 1 


1 -1 



= i - 4j - 5k 



32 Chapter 1 | Vectors 



We may also calculate the 3 x 3 determinant as 




= 2i - j - 3k - 2k - i - 3j = i - 4j - 5k. 



Areas and Volumes 

Cross products are used readily to calculate areas and volumes of certain objects. 
We illustrate the ideas involved with the next two examples. 

EXAMPLE 4 Let's use vectors to calculate the area of the triangle whose ver- 
tices are A(3, 1), B(2, —1), and C(0, 2) as shown in Figure 1.55. 













B 


Figure 1 .55 


Triangle ABC in 


Example 4. 






Figure 1 .56 Any triangle may be 
considered to be half of a 
parallelogram. 

The trick is to recognize that any triangle can be thought of as half of a 
parallelogram (see Figure 1.56) and that the area of a parallelogram is obtained 
from a cross product. In other words, A B x A C is a vector whose length measures 
the area of the parallelogram determined by and AC, and so 

Area of AABC = UlA^ x ACjl. 



To use the cross product, we must consider Tb and AC to be vectors in R 3 . 
is straightforward: We simply take the k-components to be zero. Thus, 



This 



AB = 



2j = 



2j - Ok, 



and 



Therefore, 



AC = — 3i + j = -3i + j + 0k. 



Tb 



x AC = 



j k 

-2 0 
1 0 



= -7k. 



Hence, 

Area of AABC = i|| - 7k|| = \. ♦ 



1.4 | The Cross Product 



There is nothing sacred about using A as the common vertex. We could just 
as easily have used B or C, as shown in Figure 1.57. Then 

Area of AABC = \\\b\ x B~C\\ = ~||(i-l-2j) x (-2i + 3j)|| = ±||7k|| = \. 

EXAMPLE 5 Find a formula for the volume of the parallelepiped determined 
by the vectors a, b, and c. (See Figure 1.58.) 



a x b 

B 

Figure 1.57 The area of AABC 
is 7/2. 



|c|| |cos 6\ 



a 

Figure 1 .58 The parallelepiped determined by a, b, 
and c. 

As explained in §1.3, the volume of a parallelepiped is equal to the product 
of the area of the base and the height. In Figure 1.58, the base is the parallelogram 
determined by a and b. Hence, its area is ||a x b||. The vector a x b is perpendi- 
cular to this parallelogram; the height of the parallelepiped is || c || | cos 6 1 , where 9 
is the angle between a x b and c. (The absolute value is needed in case 9 > tt/2.) 
Therefore, 



Volume of parallelepiped = (area of base)(height) 

= ||a x b|| ||c|| |cos0| 
= |(axb)-c|. 



(The appearance of the cos 9 term should alert you to the fact that dot products 
are lurking somewhere.) 

For example, the parallelepiped determined by the vectors 

a = i + 5 j , b = — 4i + 2j, and c = i + j + 6k 
has volume equal to 

|((i + 5j)x(-4i + 2j)).(i + j + 6k)| = |22k-(i + j + 6k)| 

= |22(6)| 

= 132. ♦ 

The real number (a x b) • c appearing in Example 5 is known as the triple 
scalar product of the vectors a, b, and c. Since | (a x b) • c| represents the volume 
of the parallelepiped determined by a, b, and c, it follows immediately that 





|(a x b)-c| = |(b x c)-a| = |(c x a)-b|. 



34 Chapter 1 | Vectors 




Figure 1 .59 Turning a bolt with a 
wrench. The torque on the bolt is 
the vector r x F. 




Figure 1.60 A 

potato spinning 
about an axis. 




Figure 1 .61 The 

angular velocity 
vector a>. 



In fact, if you are careful with the right-hand rule, you can convince yourself that 
the absolute value signs are not needed; that is, 

(a x b) • c = (b x c) • a = (c x a) • b. (4) 

This is a nice example of how the geometric significance of a quantity can provide 
an extremely brief proof of an algebraic property the quantity must satisfy. (Try 
proving it by writing out the expressions in terms of components to appreciate 
the value of geometric insight.) 

We leave it to you to check the following beautiful (and convenient) formula 
for calculating triple scalar products: 



(a x b) • c ■ 



ai 



(22 ^3 
C2 C 3 



where a = a\\ + ay + «3k, b = b\\ + + bj,k, and c = c\i + C2] + c^k. 

Torque 

Suppose you use a wrench to turn a bolt. What happens is the following: You 
apply some force to the end of the wrench handle farthest from the bolt and that 
causes the bolt to move in a direction perpendicular to the plane determined by 
the handle and the direction of your force (assuming such a plane exists). To 
measure exactly how much the bolt moves, we need the notion of torque (or 
twisting force). 

In particular, letting F denote the force you apply to the wrench, we have 

Amount of torque = (length of wrench)(component of F _L wrench). 

Let r be the vector from the center of the bolt head to the end of the wrench 
handle. Then 



where 9 is the ang 
torque is ||r x F||, 
as the direction in 
bolt). Hence, it is 
torque vector T is 
Note that if F 
fact that if you try 



Amount of torque = ||r|| ||F|| sin#, 

;le between r and F. (See Figure 1.59.) That is, the amount of 
and it is easy to check that the direction of r x F is the same 
which the bolt moves (assuming a right-handed thread on the 
quite natural to define the torque vector T to be r x F. The 
a concise way to capture the physics of this situation, 
is parallel to r, then T = 0. This corresponds correctly to the 
to push or pull the wrench, the bolt does not turn. 



Rotation of a Rigid Body 

Spin an object (a rigid body) about an axis as shown in Figure 1.60. What is the 
relation between the (linear) velocity of a point of the object and the rotational 
velocity? Vectors provide a good answer. 

First we need to define a vector co, the angular velocity vector of the rotation. 
This vector points along the axis of rotation, and its direction is determined by 
the right-hand rule. The magnitude of (o is the angular speed (measured in radians 
per unit time) at which the object spins. Assume that the angular speed is constant 
in this discussion. Next, fix a point O (the origin) on the axis of rotation, and let 
r(r) = OP~ be the position vector of a point P of the body, measured as a function 
of time, as in Figure 1.61. The velocity v of P is defined by 

Ar 

v = hm — , 

Ar^O At 



1.4 | The Cross Product 35 




t(0\ 









Figure 1 .62 A spinning rigid 
body. 



where Ar = r(t + At) — r(t) (i.e., the vector change in position between times t 
and t + At). Our goal is to relate v and co. 

As the body rotates, the point P (at the tip of the vector r) moves in a circle 
whose plane is perpendicular to co. (See Figure 1 .62, which depicts the motion 
of such a point of the body.) The radius of this circle is ||r(?)|| sin#, where 0 is 
the angle between co and r. Both ||r(/)|| and 6 must be constant for this rotation. 
(The direction of r(t) may change with t, however.) If At « 0, then ||Ar|| is 
approximately the length of the circular arc swept by P between / and t + At. 
That is, 

|| Ar || «s (radius of circle)(angle swept through by P) 
= (||r|| sin0)(A0) 



from the preceding remarks. Thus, 



Ar 



At 



sinO 



A<p 



Now, let At -> 0. Then Ar/Ar -> v and Acp/At 
angular velocity vector co, and we have 



co r sine 



(o x r 



\co\\ by definition of the 



(5) 




Figure 1 .63 A carousel wheel. 



It's not difficult to see intuitively that v must be perpendicular to both co and r. 
A moment's thought about the right-hand rule should enable you to establish the 
vector equation 



v = co x r. 



(6) 



If we apply formula (5) to a bicycle wheel, it tells us that the speed of a point 
on the edge of the wheel is equal to the product of the radius of the wheel and 
the angular speed (9 is n/2 in this case). Hence, if the rate of rotation is kept 
constant, a point on the rim of a large wheel goes faster than a point on the rim 
of a small one. In the case of a carousel wheel, this result tells you to sit on an 
outside horse if you want a more exciting ride. (See Figure 1.63.) 



Summary of Products Involving Vectors 

Following is a collection of some basic information concerning scalar multipli- 
cation of vectors, the dot product, and the cross product: 



Scalar Multiplication: ka 

Result is a vector in the direction of a. 
Magnitude is ||&a|| = \k\ ||a||. 
Zero if k = 0 or a = 0. 
Commutative: ka = ak. 
Associative: k(la) = (kl)a. 

Distributive: £(a + b) = ka + kb; (k + l)a = ka + la. 



Chapter 1 | Vectors 



Dot Product: a • b 




Result is a scalar 




Magnitude is a • b = Hall 1 hi cos (9* 


G is the an)?1e between a and b 


rVTapnitnde is maximized ifa II Yt 




Zero if a _L b, a = 0, or b = 0. 




Commutative: a • b = b • a. 




Associativity is irrelevant, since (a ■ 


b) • c doesn't make sense. 


Distributive: a-(b + c) = a- b + a 


• c. 


Ifa = b, then a- a = ||a|| 2 . 





Cross Product: a x b 

Result is a vector perpendicular to both a and b. 

Magnitude is ||a x b|| = ||a|| ||b|| sin#; 0 is the angle between a and b. 

Magnitude is maximized if a _L b. 

Zero ifa || b, a = 0, orb = 0. 

Anticommutative: axb = — bxa. 

Not associative: In general, a x (b x c) ^ (a x b) x c. 

Distributive: ax(b + c) = axb + axc and 
(a + b)xc = axc + bxc. 

Ifa _Lb, then ||axb|| = ||a|| ||b||. 



Addendum: Proofs of Cross Product Properties 

Proof of Property 1 To prove the anticommutativity property, we use the right- 
hand rule. Since 

||axb|| = ||a|| ||b|| sin0, 

we obviously have that ||a x b|| = ||b x a||. Therefore, we need only understand 
the relation between the direction of a x b and that of b x a. To determine the 
direction of a x b, imagine curling the fingers of your right hand from a toward b. 
Then your thumb points in the direction of a x b. If instead you curl your fingers 
from b toward a, then your thumb will point in the opposite direction. This is the 
direction of b x a, so we conclude that a x b = — b x a. (See Figure 1.64.) ■ 



ax b 




Figure 1 .64 The right-hand rule shows why a x b = — b x a. 



1.4 | The Cross Product 37 
Proof of Property 2 First, note the following general fact: 



PROPOSITION 4.3 Let a and b be vectors in R 3 . If a • x = b • x for all vectors x 
in R 3 , then a = b. 



To establish Proposition 4.3, write a as aii + ci2] + 03k and b as b\i + + 
b^k and set x in turn equal to i, j, and k. Proposition 4.3 is valid for vectors in R 2 
as well as R 3 . 

To prove the distributive law for cross products (property 2), we show that, 
for any x e R 3 , 

(a x (b + c)) • x = (a x b + a x c) • x. 

By Proposition 4.3, property 2 follows. 
From the equations in (4), 

(a x (b + c)) • x = (x x a) • (b + c) 

= (x x a) • b + (x x a) • c, 

from the distributive law for dot products, 

= (a x b) • x + (a x c) • x 
= (a x b + a x c) • x, 

again using (4) and the distributive law for dot products. ■ 

Proof of Property 3 Property 3 follows from properties 1 and 2. We leave the 
details as an exercise. ■ 

Proof of Property 4 The second equality in property 4 follows from the first 
equality and property 1 : 

k(a x b) = —k(b x a) by property 1 

= —(kb) x a by the first equality of property 4 
= a x (kb) by property 1 . 

Hence, we need only prove the first equality. 

If either a or b is the zero vector or if a is parallel to b, then the first equality 
clearly holds. Otherwise, we divide into three cases: (1) k = 0, (2) k > 0, and 
(3) k < 0. If k = 0, then both ka and k(a x b) are equal to the zero vector and 
the desired result holds. If k > 0, the direction of (ka) x b is the same as a x b, 
which is also the same as k(a x b). Moreover, the angle between ka and b is the 
same as between a and b. Calling this angle 0, we check that 

||(*a)xb|| = ||*a|| ||b|| sine 

= &||a|| ||b|| sin# by part 1 of Proposition 3.4 

= k || a x b || by Definition 4. 1 

= \\k(a x b)|| by part 1 of Proposition 3.4. 

We conclude (ka) x b = k(a x b) in this case. 



38 Chapter 1 | Vectors 




a 



fca, k < 0 



fca, k > 0 



Figure 1 .65 If the angle between 
a and b is 0, then the angle 
between £a and b is either 0 (if 
k > 0) or?r -0 (if£ < 0). 



If k < 0, then the direction of (ka) x b is the same as that of (—a) x b, which 
is seen to be the same as that of —(a x b) and thus the same as that of k(a x b). 
The angle between ka and b is therefore n — 9, where 9 is the angle between a 
and b. (See Figure 1.65.) Thus, 

||(*a) x b|| = ||&a|| ||b|| sin(7r - 9) = \k\ ||a|| ||b|| sin 9 = ||*(a x b)||. 

So, again, it follows that (ka) x b = k(a x b). ■ 



1.4 Exercises 



Evaluate the determinants in Exercises 1—4. 



1. 



3. 



2. 



4. 



In Exercises 5—7, calculate the indicated cross products, using 
both formulas (2) and (3). 

5. (1, 3, -2) x (-1,5, 7) 

6. (3i-2j + k)x(i + j + k) 

7. (i + j)x(-3i + 2j) 

8. Prove property 3 of cross products, using properties 1 
and 2. 

9. If a x b = 3i - 7j - 2k, what is (a + b) x (a - b)? 

1 0. Calculate the area of the parallelogram having vertices 
(1,1), (3, 2), (1,3), and (-1,2). 

1 1 . Calculate the area of the parallelogram having vertices 
(1,2, 3), (4, -2, 1), (-3, 1, 0), and (0, -3, -2). 

1 2. Find a unit vector that is perpendicular to both 2i + 
j — 3k and i + k. 

13. If (a x b) • c = 0, what can you say about the geomet- 
ric relation between a, b, and c? 

Compute the area of the triangles described in Exercises 
14-17. 

1 4. The triangle determined by the vectors a = i + j and 
b = 2i-j 

15. The triangle determined by the vectors a = i — 2j + 
6k and b = 4i + 3j - k 

16. The triangle having vertices (1,1), (—1,2), and 
(-2,-1) 

17. The triangle having vertices (1,0, 1), (0,2,3), and 
(-1,5,-2) 



1 8. Find the volume of the parallelepiped determined by 
a = 3i - j, b = -2i + k, and c = i - 2j + 4k. 

1 9. What is the volume of the parallelepiped with vertices 
(3,0,-1), (4,2,-1), (-1,1,0), (3,1,5), (0,3,0), 
(4,3,5), (-1,2, 6), and (0,4,6)? 



20. Verify that (a x b) • c 



a i 



a 2 a-} 
b 2 b 3 
C2 c 3 



21 . Show that (a x b) • c = a • (b x c) using Exercise 20. 

22. Use geometry to show that |(axb)-c| = 
|b-(axc)|. 

23. (a) Show that the area of the triangle with vertices 

P\(xi,y\), P 2 (x 2 , yi), and P 3 (x 3 , y 3 ) is given by 
the absolute value of the expression 



1 1 1 

x\ x 2 JC 3 

yi yi yi 



(b) Use part (a) to find the area of the triangle with 
vertices (1,2), (2, 3), and (-4, -4). 

24. Suppose that a, b, and c are noncoplanar vectors in R 3 , 
so that they determine a tetrahedron as in Figure 1 .66. 




Figure 1.66 The tetrahedron of 
Exercise 24. 

Give a formula for the surface area of the tetrahedron 
in terms of a, b, and c. (Note: More than one formula 
is possible.) 



1.4 | Exercises 39 



25. Suppose that you are given nonzero vectors a, b, and c 
in R 3 . Use dot and cross products to give expressions for 
vectors satisfying the following geometric descriptions: 

(a) A vector orthogonal to a and b 

(b) A vector of length 2 orthogonal to a and b 

(c) The vector projection of b onto a 

(d) A vector with the length of b and the direction of a 

(e) A vector orthogonal to a and b x c 

(f) A vector in the plane determined by a and b and 
perpendicular to c. 

26. Suppose a, b, c, and d are vectors in R 3 . Indicate which 
of the following expressions are vectors, which are 
scalars, and which are nonsense (i.e., neither a vector 
nor a scalar). 



(a) (a x b) x c 

(c) (a • b) x (c • d) 

(e) (a • b) x (c x d) 

(g) (a x b) • (c x d) 



(b) (a-b)-c 
(d) (a x b) • c 
(f) ax[(b-c)d] 
(h) (a • b)c - (a x b) 



Exercises 27—32 concern several identities for vectors a, b, c, 
and d in R 3 . Each of them can be verified by hand by writing 
the vectors in terms of their components and by using formula 
(2) for the cross product and Definition 3.1 for the dot product. 
However, this is quite tedious to do. Instead, use a computer 
algebra system to define the vectors a, b, C, and d in general 
and to verify the identities. 

^27. (a x b) x c = (a • c)b - (b • c)a 

^ 28. a • (b x c) = b • (c x a) = c • (a x b) 

= -a • (c x b) = -c • (b x a) 
= -b • (a x c) 

^ 29. (a x b) • (c x d) = (a • c)(b • d) - (a • d)(b • c) 
a • c a • d 
b-c b-d 

^ 30. (a x b) x c + (b x c) x a + (c x a) x b = 0 (this is 
known as the Jacobi identity). 

^ 31. (a x b) x (c x d) = [a • (c x d)]b - [b • (c x d)]a 
^ 32. (a x b) • (b x c) x (c x a) = [a • (b x c)] 2 

33. Establish the identity 

(a x b) • (c x d) = (a • c)(b • d) - (a • d)(b • c) 

of Exercise 29 without resorting to a computer algebra 
system by using the results of Exercises 27 and 28. 

34. Egbert applies a 20 lb force at the edge of a 4 ft 
wide door that is half-open in order to close it. (See 
Figure 1.67.) Assume that the direction of force is per- 
pendicular to the plane of the doorway. What is the 
torque about the hinge on the door? 

35. Gertrude is changing a flat tire with a tire iron. The tire 
iron is positioned on one of the bolts of the wheel so 




Figure 1.67 Figure for Exercise 34. 




40 lb 



Figure 1.68 The configuration for 
Exercise 35. 

that it makes an angle of 30° with the horizontal. (See 
Figure 1.68.) Gertrude exerts 40 lb of force straight 
down to turn the bolt. 

(a) If the length of the arm of the wrench is 1 ft, how 
much torque does Gertrude impart to the bolt? 

(b) What if she has a second tire iron whose length is 
18 in? 

36. Egbert is trying to open a jar of grape jelly. The ra- 
dius of the lid of the jar is 2 in. If Egbert imparts 15 lb 
of force tangent to the edge of the lid to open the jar, 
how many ft-lb, and in what direction, is the resulting 
torque? 

37. A 50 lb child is sitting on one end of a seesaw, 3 ft 
from the center fulcrum. (See Figure 1 .69.) When she is 




1.5 ft 



Figure 1.69 The seesaw of Exercise 37. 



40 Chapter 1 | Vectors 



1 .5 ft above the horizontal position, what is the amount 
of torque she exerts on the seesaw? 

38. For this problem, note that the radius of the earth is 
approximately 3960 miles. 

(a) Suppose that you are standing at 45° north latitude. 
Given that the earth spins about its axis, how fast 
are you moving? 

(b) How fast would you be traveling if, instead, you 
were standing at a point on the equator? 

39. Archie, the cockroach, and Annie, the ant, are on an 
LP record. Archie is at the edge of the record (ap- 
proximately 6 in from the center) and Annie is 2 in 
closer to the center of the record. How much faster is 
Archie traveling than Annie? (Note: A record playing 
on a turntable spins at a rate of 33 j revolutions per 
minute.) 

40. A top is spinning with a constant angular speed of 12 
radians/sec. Suppose that the top spins about its axis 



of symmetry and we orient things so that this axis is 
the z-axis and the top spins counterclockwise about it. 

(a) If, at a certain instant, a point P in the top has 
coordinates (2,-1, 3), what is the velocity of the 
point at that instant? 

(b) What are the (approximate) coordinates of P one 
second later? 

41. There is a difficulty involved with our definition of 
the angular velocity vector go, namely, that we cannot 
properly consider this vector to be "free" in the sense 
of being able to parallel translate it at will. Consider 
the rotations of a rigid body about each of two parallel 
axes. Then the corresponding angular velocity vectors 
go 1 and <w 2 are parallel. Explain, perhaps with a fig- 
ure, that even if G0 l and G0 2 are equal as "free vectors," 
the corresponding rotational motions that result must 
be different. (Therefore, when considering more than 
one angular velocity, we should always assume that the 
axes of rotation pass through a common point.) 




Figure 1 .70 The plane in R 3 
through the point Pq and 
perpendicular to the vector n. 



1 .5 Equations for Planes; Distance Problems 

In this section, we use vectors to derive analytic descriptions of planes in R 3 . We 
also show how to solve a variety of distance problems involving "fiat objects" 
(i.e., points, lines, and planes). 

Coordinate Equations of Planes 

A plane IT in R 3 is determined uniquely by the following geometric information: 
a particular point Pq(xo, vo, zo) in the plane and a particular vector n = Ai + 
B\ + Ck that is normal (perpendicular) to the plane. In other words, n is the 

set of all points P(x, y, z) in space such that PqP is perpendicular to n. (See 
Figure 1.70.) This means that n is defined by the vector equation 



n- Kf> = 0. 



(1) 



Since PqP = (x — xo)i + (y — yo)j + (z — zo)k, equation (1) may be rewritten 
as 

(Ai + fij + Ck) • ((x - x 0 )i + (y - y 0 )j + (z - z 0 )k) = 0 

or 



A(x - x 0 ) + B(y - y 0 ) + C(z - z 0 ) = 0. 



(2) 



This is equivalent to 

Ax + By + Cz = D, 
where D = Axq + Byo + Czo- 



1.5 | Equations for Planes; Distance Problems 41 



EXAMPLE 1 Theplanethroughthepoint(3,2, 1) with normal vector 2i — j + 4k 
has equation 

(2i - j + 4k) • ((x - 3)i + (y - 2)j + ( z - l)k) = 0 
<=► 2(x - 3) - (y - 2) + 4(z - 1) = 0 

2x - y + 4z = 8. # 

Not only does a plane in R 3 have an equation of the form given by equation 
(2), but, conversely, any equation of this form must describe a plane. Moreover, 
it is easy to read off the components of a vector normal to the plane from such an 
equation: They are just the coefficients of x, y, and z. 

EXAMPLE 2 Given the plane with equation Ix + 2y — 3z = 1, find a normal 
vector to the plane and identify three points that lie on that plane. 

A possible normal vector is n = 7i + 2j — 3k. However, any nonzero scalar 
multiple of n will do just as well. Algebraically, the effect of using a scalar multiple 
of n as normal is to multiply equation (2) by such a scalar. 

Finding three points in the plane is not difficult. First, let y = z = 0 in the 
defining equation and solve for x: 

Ix + 2-0-3-0=1 <^=> 7x = l x=f 

Thus (j, 0, 0) is a point on the plane. Next, let x = z = 0 and solve for y: 

7-0 + 2^-3-0=1 y=\. 

So (0, j, 0) is another point on the plane. Finally, let x = y = 0 and solve for z. 
You should find that (0,0,— j) lies on the plane. ♦ 

EXAMPLE 3 Put coordinate axes on R 3 so that the z-axis points vertically. 
Then a plane in R 3 is vertical if its normal vector n is horizontal (i.e., if n is 
parallel to the xy-plane). This means that n has no k-component, so n can be 
written in the form Ai + Bj. It follows from equation (2) that a vertical plane has 
an equation of the form 

A(x - x 0 ) + B(y - yo) = 0. 
Hence, a nonvertical plane has an equation of the form 

A(x - x 0 ) + B(y - y 0 ) + C(z - z 0 ) = 0, 
where C / 0. ♦ 

EXAMPLE 4 From high school geometry, you may recall that a plane is 
determined by three (noncollinear) points. Let's find an equation of the plane 
that contains the points P 0 (l, 2, 0), Pi(3, 1, 2), and P 2 (0, 1, 1). 

There are two ways to solve this problem. The first approach is algebraic 
and rather uninspired. From the aforementioned remarks, any plane must have 
an equation of the form Ax + By + Cz = D for suitable constants A, B, C, and 
D. Thus, we need only to substitute the coordinates of Po, Pi, and P2 into this 
equation and solve for A, B, C, and D. We have that 

• substitution of Po gives A + 2B = D; 

• substitution of Pi gives 3 A + B + 2C = D; and 

• substitution of P2 gives B + C = D. 



42 Chapter 1 | Vectors 




Figure 1.71 The plane 
determined by the points Pq, P\, 
and P2 in Example 4. 



Hence, we must solve a system of three equations in four unknowns: 

A + 2B = D 

3 A + B + 2C = D . 
B + C = D 



(3) 



In general, such a system has either no solution or else infinitely many solutions. 
We must be in the latter case, since we know that the three points Pq, P\, and P2 
lie on some plane (i.e., that some set of constants A, B, C, and D must exist). 
Furthermore, the existence of infinitely many solutions corresponds to the fact 
that any particular equation for a plane may be multiplied by a nonzero constant 
without altering the plane defined. In other words, we can choose a value for one 
of A, B, C, or D, and then the other values will be determined. So let's multiply 
the first equation given in (3) by 3, and subtract it from the second equation. We 
obtain 



A+ 2B = D 

-5B +2C = -2D . 
B + C = D 



(4) 



Now, multiply the third equation in (4) by 5 and add it to the second: 

A + 2B = D 

1C = 3D . (5) 
B + C = D 

Multiply the third equation appearing in (5) by 2 and subtract it from the first: 

A -2C = -D 

1C= 3D . (6) 
B + C = D 

By adding appropriate multiples of the second equation to both the first and third 
equations of (6), we find that 



1C = 



3D 



(7) 



B 



= %D 



Thus, if in (7) we take D = —1 (for example), then A = 1, B = —4, C = —3, 
and the equation of the desired plane is 

x - 4y - 3z = -7. 

The second method of solution is cleaner and more geometric. The idea is 
to make use of equation (1). Therefore, we need to know the coordinates of a 
particular point on the plane (no problem — we are given three such points) and 

a vector n normal to the plane. The vectors P0P1 and P0P2 both lie in the plane. 
(See Figure 1.71.) In particular, the normal vector n must be perpendicular to 
them both. Consequently, the cross product provides just what we need. That is, 
we may take 



n = P 0 Pi x P 0 P 2 = (2i - j + 2k) x (-i - j + k) 



= i - 4j - 3k. 



1.5 | Equations for Planes; Distance Problems 43 




2y + z = 4 



Figure 1 .72 The line of 

intersection of the planes 

x — 2y + z = 4 and 

2x + y + 3z = —7 in Example 5. 



If we take Po(l, 2, 0) to be the particular point in equation (1), we find that the 
equation we desire is 

(i-4j-3k)-((x- l)i + (v-2)j + zk) = 0 

or 

(x - l) - My - 2) - 3z = 0. 
This is the same equation as the one given by the first method. ♦ 

EXAMPLE 5 Consider the two planes having equations x — 2 y + z = 4 and 

2x + y + 3z = — 7. We determine a set of parametric equations for their line of 
intersection. (See Figure 1.72.) We use Proposition 2.1. Thus, we need to find a 
point on the line and a vector parallel to the line. To find the point on the line, 
we note that the coordinates (x, y, z) of any such point must satisfy the system of 
simultaneous equations given by the two planes 



x-2y + z= 4 
2x + y + 3z = -7 



(8) 



From the equations given in (8), it is not too difficult to produce a single 
solution (x, y, z). For example, if we let z = 0 in (8), we obtain the simpler 
system 




(9) 
-2, 



The solution to the system of equations (9) is readily calculated to be x = 
y = —3. Thus, (—2, —3, 0) are the coordinates of a point on the line. 

To find a vector parallel to the line of intersection, note that such a vector 
must be perpendicular to the two normal vectors to the planes. The normal vectors 
to the planes are i — 2j + k and 2i + j + 3k. Therefore, a vector parallel to the 
line of intersection is given by 

(i - 2j + k) x (2i + j + 3k) = -7i - j + 5k. 

Hence, Proposition 2.1 implies that a vector parametric equation for the line is 

r(r) = (-2i - 3j) + f(-7i - j + 5k), 

and a standard set of parametric equations is 

x = -7f - 2 
y = -t-3 . 

z = 5t ♦ 



Parametric Equations of Planes 

Another way to describe a plane in R 3 is by a set of parametric equations. First, 
suppose that a = (ct\, a-i, a^) and b = (b\, b 2 , b^) are two nonzero, nonparallel 
vectors in R 3 . Then a and b determine a plane in R 3 that passes through the 
origin. (See Figure 1.73.) To find the coordinates of a point P(x, y, z) in this 
plane, draw a parallelogram whose sides are parallel to a and b and that has two 
opposite vertices at the origin and at P, as shown in Figure 1 .74. Then there must 
exist scalars s and t so that the position vector of P is ssi + tb. The plane 



44 Chapter 1 | Vectors 



z 




Figure 1 .73 The plane through 
the origin determined by the 
vectors a and b. 



z 




y 



Figure 1 .74 For the point P in 
the plane shown, OP = .?a + fb 
for appropriate scalars s and t. 




Figure 1 .75 The plane passing 
through Pq(c\ , C2, C3) and parallel 
to a and b. 



may be described as 

{x e R 3 | x = sa + fb; s, t e R} . 

Now, suppose that we seek to describe a general plane n (i.e., one that does 
not necessarily pass through the origin). Let 

c = (ci, c 2 , c 3 ) = OPq 

denote the position vector of a particular point P 0 in n and let a and b be two 
(nonzero, nonparallel) vectors that determine the plane through the origin parallel 
to n. By parallel translating a and b so that their tails are at the head of c (as in 
Figure 1.75), we adapt the preceding discussion to see that the position vector of 
any point P(x, y, z) in n may be described as 

= sa + tb + c. 
To summarize, we have shown the following: 

PROPOSITION 5.1 A vector parametric equation for the plane n containing 
the point Pq(c\, c 2 , c?,) (whose position vector is OPq = c) and parallel to the 
nonzero, nonparallel vectors a and b is 

x(s, t) = sa+tb + c. (10) 



By taking components in formula (10), we readily obtain a set of parametric 
equations for n : 



x = sa\ + tb\ + C\ 

y = sa 2 + tb 2 + c 2 ■ (11) 
Z = scii + tb-i + c 3 



Compare formula (10) with that of equation (1) in Proposition 2.1. We need 
to use two parameters s and t to describe a plane (instead of a single parameter 
t that appears in the vector parametric equation for a line) because a plane is a 
two-dimensional object. 



1.5 | Equations for Planes; Distance Problems 45 



EXAMPLE 6 We find a set of parametric equations for the plane that passes 
through the point (1,0,-1) and is parallel to the vectors 3i — k and 2i + 5j + 2k. 
From formula (10), any point on the plane is specified by 



x(s, t) = s(3i - k) + r(2i + 5j + 2k) + (i 
= (3s + 2t + l)i + 5t\ + (2/ - s - 
The individual parametric equation may be read off as 

x = 3s + 2t + 1 



-k) 
l)k. 



y = 5t 
z = 2t 



1 




Figure 1.76 A general 
configuration for finding the 
distance between a point and a 
line, using vector projections. 




Figure 1 .77 Another general 
configuration for finding the 
distance between a point and a 
line. 



Distance Problems 

Cross products and vector projections provide convenient ways to understand a 
range of distance problems involving lines and planes: Several examples follow. 
What is important about these examples are the vector techniques for solving 
geometric problems that they exhibit, not the general formulas that may be derived 
from them. 

EXAMPLE 7 (Distance between a point and a line) We find the distance 
between the point P 0 (2, 1 , 3) and the line 1(0 = t(- 1 , 1 , -2) + (2, 3 , -2) in two 
ways. 

Method 1. From the vector parametric equations for the given line, we read 
off a point B on the line — namely, (2, 3, —2) — and a vector a parallel to the 
line — namely, a = (— 1, 1,-2). Using Figure 1.76, the length of the vector 
BPq — proj a SPo provides the desired distance between P 0 and the line. Thus, 
we calculate that 

B~?q = {2, 1,3) -(2, 3,-2) 
= (0, -2, 5); 



proj a BP 0 = 



a- BP, 



o 



a • a 



(-1, 1,-2). (0,-2, 5) 



(-1,1,-2) 



" V("l,l,-2)-(-l,l,-2X 
= (2, -2, 4). 

The desired distance is 

||BP 0 -proj a fiP 0 || = 11(0, -2, 5) -(2, -2, 4)|| = ||(-2, 0, 

Method 2. In this case, we use a little trigonometry. If 0 denotes the angle 
between the vectors a and BPq as in Figure 1.77, then 

D 

sin 6* = 



= V~5. 



\\BP 0 \ 



where D denotes the distance between Pq and the line. Hence, 

|| a || || Mil sine HaxBPol 



D= \\BP 0 \\ sin<9 = 



a 



46 Chapter 1 | Vectors 



Therefore, we calculate 

a x B~fi) = 

so that the distance sought is 
D = 



i j k 

-1 1 -2 
0-2 5 



|i + 5j + 2k|| 



-i + j-2k|| 76 
which agrees with the answer obtained by Method 1 . 



= i + 5j + 2k, 




Figure 1 .78 The general 
configuration for finding the 
distance D between two parallel 
planes. 








/CJL — c 





Bo 



Figure 1.79 Configuration for 
determining the distance between 
two skew lines in Example 9. 



EXAMPLE 8 (Distance between parallel planes) The planes 
111: 2x -2y+z = 5 and n 2 : 2x - 2y + z = 20 

are parallel. (Why?) We see how to compute the distance between them. 

Using Figure 1.78 as a guide, we see that the desired distance D is given by 
||proj n / , ii D 2||, where Pi is a point on Oi, P 2 is a point on n 2 , and n is a vector 
normal to both planes. 

First, the vector n that is normal to both planes may be read directly from 
the equation for either n 1 or Tl2 as n = 2i — 2j + k. It is not hard to find a point 
Pi on TI 1 : the point Pi(0, 0, 5) will do. Similarly, take P 2 (0, 0, 20) for a point on 
n 2 . Then 



P,P 2 = (0,0, 15), 



and calculate 



proj n P 1 P 2 = 



n-PiP 2 



n • n 



n = 



(2,-2, 1). (0,0, 15) 



(2. 



= -¥(2, 



-2,1) -(2, 

v-, "2, 1) 
= -§(2,-2,1). 
Hence, the distance D that we seek is 
D 



-2, 1) 



(2,-2,1) 



llproj n Pi/> 2 



\V9 



EXAMPLE 9 (Distance between two skew lines) Find the distance between 
the two skew lines 

li(0 = *(2. 1,3) + (0, 5,-1) and l 2 (r) = f(l, -1, 0) + (-1, 2, 0). 

(Two lines in R 3 are said to be skew if they are neither intersecting nor parallel. 
It follows that the lines must lie in parallel planes and that the distance between 
the lines is equal to the distance between the planes.) 

To solve this problem, we need to find ||proj n 5i B 2 1| , the length of the projec- 
tion of the vector between a point on each line onto a vector n that is perpendicular 
to both lines, hence, also perpendicular to the parallel planes that contain the lines. 
(See Figure 1.79.) 

From the vector parametric equations for the lines, we read that the point 
Bi(0, 5, —1) is on the first line and 5 2 (— 1, 2, 0) is on the second. Hence, 



B,B 2 = (-1, 2, 0) - (0, 5, -1) = (-1, -3, 1). 



1.5 | Exercises 47 



For a vector n that is perpendicular to both lines, we may use n = ai x a 2 , where 
ai = (2, 1 , 3) is a vector parallel to the first line and a2 = ( 1 , — 1 , 0) is parallel to 
the second. (We may read these vectors from the parametric equations.) Thus, 



n = ai x a2 



and so, 



proj n £i£ 2 



n- BiB 2 



n • n 



J 



= 3i + 3j - 3k, 



(-1,-3, l)-(3,3,-3) 
(3,3, -3) -(3, 3, -3) 



(3,3,-3) 



(3,3, 



= -f(l,l, 



-3) 
!)• 



The desired distance is ||proj nJ Si Z?2 II = |V3. 



1.5 Exercises 



1 . Calculate an equation for the plane containing the point 
(3,-1,2) and perpendicular to i — j + 2k. 

2. Find an equation for the plane containing the point 
(9, 5, —1) and perpendicular to i — 2k. 

3. Find an equation for the plane containing the points 
(3,-1,2), (2, 0,5), and (1, -2,4). 

4. Find an equation for the plane containing the points 
(A, 0, 0), (0, B, 0), and (0, 0, C). Assume that at least 
two of A, B, and C are nonzero. 

5. Give an equation for the plane that is parallel to the 
plane 5x — 4y + z = 1 and that passes through the 
point (2, —1, —2). 

6. Give an equation for the plane parallel to the plane 2x — 
3y + z = 5 that passes through the point (— 1 , 1,2). 

7. Find an equation for the plane parallel to the plane x — 
y + Iz = 10 that passes through the point (—2, 0, 1). 

8. Give an equation for the plane parallel to the plane 
2x + 2y + z = 5 and that contains the line with para- 
metric equations x = 2 — t, y = 2t + 1, z = 3 — 2t. 

9. Explain why there is no plane parallel to the plane 
5x — 3 y + 2z = 10 that contains the line with para- 
metric equations x = t + 4, y = 3t — 2, z = 5 — 2t . 

1 0. Find an equation for the plane that contains the line x = 
2t - 1, y = 3t + 4, z = 1 - t and the point (2, 5, 0). 

11. Find an equation for the plane that is perpendicular 
to the line x = It - 5, y = 7 - It, z = 8 - t and that 
passes through the point (1, —1,2). 



12. Find an equation for the plane that contains the two 
lines l\.x = t + 2, y = 3t — 5,z = 5t + 1 andh'.x = 

5 - t, y = 3t - 10, z = 9 - 2t. 

1 3. Give a set of parametric equations for the line of inter- 
section of the planes x + 2y — 3z = 5 and 5x + 5y — 
z = 1. 

1 4. Give a set of parametric equations for the line through 
(5, 0, 6) that is perpendicular to the plane 2x — 3y + 
5z = -1. 

1 5. Find a value for A so that the planes Sx — 6y + 9 Az = 

6 and Ax + y + 2z = 3 are parallel. 

16. Find values for A so that the planes Ax — y + z = 1 
and 3 Ax ; + Ay — 2z = 5 are perpendicular. 

Give a set of parametric equations for each of the planes de- 
scribed in Exercises 1 7—22. 

17. The plane that passes through the point (—1, 2, 7) and 
is parallel to the vectors 2i — 3j + k and i — 5k 

18. The plane that passes through the point (2,9,-4) 
and is parallel to the vectors — 8i + 2j + 5k and 
3i - 4j - 2k 

19. The plane that contains the lines h: x = 2t + 5, y = 

-3t - 6,z = 4f + 10 and l 2 : x = 5t - 1, y = lOt + 
3,z = lt-2 

20. The plane that passes through the three points (0, 2, 1), 
(7, -1,5), and (-1,3,0) 

21. The plane that contains the line /: x = 3t — 5, y = 
10 - 3t, z = 2t + 9 and the point (-2, 4, 7) 



48 Chapter 1 | Vectors 



22. The plane determined by the equation 2x — 3y + 
5z = 30 

23. Find a single equation of the form Ax + By + Cz = D 
that describes the plane given parametrically as x = 
3s - t + 2, y = 4s + t, z = s + 5t + 3. (Hint: Begin 
by writing the parametric equations in vector form and 
then find a vector normal to the plane.) 

24. Find the distance between the point (1 , —2, 3) and the 
line I: x = 2t - 5, y = 3 - t, z = 4. 

25. Find the distance between the point (2, —1) and the 

line/:jc = 3f + 7, y = 5t - 3. 

26. Find the distance between the point (—11, 10, 20) and 
the line /: x = 5 - t , y = 3, z = It + 8. 

27. Determine the distance between the two lines li(f) = 
f(8, -1,0) + (-1,3, 5) and I 2 (f) = f(0, 3, 1) + 
(0, 3, 4). 

28. Compute the distance between the two lines 
h(0 = (t ~ 7)i +(5f + l)j + (3 -2f)k and l 2 (r) = 
4ri + (2-0j + (8f+l)k. 

29. (a) Find the distance between the two lines li(f) = 

f(3, 1,2) + (4, 0,2) and I 2 (f) = t(\, 2, 3) + 
(2,1,3). 

(b) What does your answer in part (a) tell you about 
the relative positions of the lines? 

30. (a) The lines h(r) = t(l, -1, 5) + (2, 0, -4) and 

l 2 (r) = ?(l,-l,5) + (l,3,-5) are parallel. Ex- 
plain why the method of Example 9 cannot be used 
to calculate the distance between the lines. 

(b) Find another way to calculate the distance. (Hint: 
Try using some calculus.) 

31. Find the distance between the two planes given by the 
equations x — 3 y + 2z = 1 and x — 3y + 2z = 8. 

32. Calculate the distance between the two planes 

5x - 2y + 2z = 12 and - 10.v + 4y - 4z = 8. 

33. Show that the distance d between the two parallel 
planes determined by the equations Ax + By + Cz = 



Di and Ax + By + Cz = D 2 is 
d= \Di-D 2 \ 

Va 2 + b 2 + c 2 ' 

34. Two planes are given parametrically by the vector 
equations 

Xl (s, f) = (-3, 4, -9) + s(9, -5, 9) + f(3, -2, 3) 
x 2 (s, t) = (5, 0, 3) + s(-9, 2, -9) + r(-4, 7, -4). 

(a) Give a convincing explanation for why these 
planes are parallel. 

(b) Find the distance between the planes. 

35. Write equations for the planes that are parallel to 
x + 3y — 5z = 2 and lie three units from it. 

36. Suppose that li(r) = fa + bi and l 2 (f) = ra + b 2 are 
parallel lines in either R 2 or R 3 . Show that the distance 
D between them is given by 

D= ||ax(b 2 -bi)|| 
11*11 

(Hint: Consider Example 7.) 

37. Let n be the plane in R 3 with normal vector n that 
passes through the point A with position vector a. If b 
is the position vector of a point B in R 3 , show that the 
distance D between B and n is given by 

|n-(b-a)| 

D = -. 

1 1 n 1 1 

38. Show that the distance D between parallel planes with 
normal vector n is given by 

|n-(x 2 -xQ| 
1 1 n 1 1 

where X] is the position vector of a point on one of the 
planes, and x 2 is the position vector of a point on the 
other plane. 

39. Suppose that li (f) = t&i + bi andl 2 (f) = fa 2 + b 2 are 
skew lines in R 3 . Use the geometric reasoning of Ex- 
ample 9 to show that the distance D between these lines 
is given by 

_ |(a t x a 2 )-(b 2 -bQ| 
|| ai x a 2 || 



1 .6 Some n-dimensional Geometry 

Vectors in R" 

The algebraic idea of a vector in R 2 or R 3 is defined in § 1 . 1 , in which we asked 
you to consider what would be involved in generalizing the operations of vector 
addition, scalar multiplication, etc., to n-dimensional vectors, where n can be 
arbitrary. We explore some of the details of such a generalization next. 



1.6 | Some n-dimensional Geometry 49 



DEFINITION 6.1 A vector in R" is an ordered rc-tuple of real numbers. We 
use a = (til, a 2 , • • ■ , a n ) as our standard notation for a vector in R". 



EXAMPLE 1 The 5-tuple (2, 4, 6, 8, 10) is a vector in R 5 . The (n + l)-tuple 
(2«, 2rc — 2, 2« — 4, . . . , 2, 0) is a vector in R" +1 , where « is arbitrary. ♦ 

Exactly as is the case in R 2 or R 3 , we call two vectors a = (a\ , a 2 , . . . , a n ) and 

b = (pi, b 2 Z?„) equal just incases, = b t for/ = 1,2,..., n. Vector addition 

and scalar multiplication are defined in complete analogy with Definitions 1 .3 and 
1.4: If a = (a\, a 2 , . . . , a n ) and b = (b\ , b 2 , . . . , b„) are two vectors in R" and 
k e R is any scalar, then 

a + b = (ai + b\, a 2 + b 2 a n + fc„) 

and 

k& = (ka\, ka 2 , . . . , ^a n ). 

The properties of vector addition and scalar multiplication given in §1.1 hold 
(with proofs that are no different from those in the two- and three-dimensional 
cases). Similarly, the dot product of two vectors in R" is readily defined: 

a • b = fli&i + a 2 b 2 + • • • + a„b n . 

The dot product properties given in §1.3 continue to hold in n dimensions; we 
leave it to you to check that this is so. 

What we cannot do in dimensions larger than three is to develop a pictorial 
representation for vectors as arrows. Nonetheless, the power of our algebra and 
analogy does allow us to define a number of geometric ideas. We define the length 
of a vector in a e R" by using the dot product: 

||a|| = Va^a. 

The distance between two vectors a and b in R" is 

Distance between a and b = ||a — b||. 

We can even define the angle between two nonzero vectors by using a generalized 
version of equation (4) of § 1 .3 : 



Here a, be R" and 9 is taken so that 0 < 6 < it. (Note: At this point in our 
discussion, it is not clear that we have 

a-b 

-1 < < 1, 

" Ila|| ||b|| " 

which is a necessary condition if our definition of the angle 9 is to make sense. 
Fortunately, the Cauchy-Schwarz inequality — formula (1) that follows — takes 
care of this issue.) Thus, even though we are not able to draw pictures of vectors 
in R", we can nonetheless talk about what it means to say that two vectors are 
perpendicular or parallel, or how far apart two vectors may be. (Be careful about 
this business. We are defining notions of length, distance, and angle entirely in 



Chapter 1 | Vectors 



terms of the dot product. Results like Theorem 3.3 have no meaning in R", since 
the ideas of angles between vectors and dot products are not independent.) 

There is no simple generalization of the cross product. However, see Exer- 
cises 39^12 at the end of this section for the best we can do by way of analogy. 

We can create a standard basis of vectors in R" that generalize the i, j, 
k-basis in R 3 . Let 

d =(1,0, 0.....0), 
e 2 =(0, 1,0,..., 0), 



e„ = (0, 0, . . . , 0, 1). 
Then it is not difficult to see (check for yourself) that 

a = (ai, a 2 , . . . , a„) = a\&\ + a 2 e 2 H h a„e„ 

Here are two famous (and often handy) inequalities: 




PROOF If n = 2 or 3, this result is virtually immediate in view of Theorem 3.3. 
However, in dimensions larger than three, we do not have independent notions of 
inner products and angles, so a different proof is required. 

First note that the inequality holds if either a or b is 0. So assume that a and 
b are nonzero. Then we may define the projection of b onto a just as in §1.3: 

/a-b\ 
projab = I — I a = ka. 

Here k is, of course, the scalar a • b/a • a. Let c = b — ka (so that b = ka + c). 
Then we have a • c = 0, since 



a • c 



a • a 



= a 


(b- 


- jfca) 


= a 


b- 


ka • a 
/a-b 


= a 


b- 


\a • a 


= a 


b- 


a-b 


= 0. 







We leave it to you to check that the "Pythagorean theorem" holds, namely, that 
the following equation is true: 



|b|| 2 = £ 2 ||a|| 2 + ||c|| 



Multiply this equation by || a 

2 



\2 - 



a • a. We obtain 



a 



b|| 2 = 



|a|| 2 £ 2 ||a|| 2 + ||a|| 2 ||c|| 2 



= a 



a-b 



a • a 



|a|| 2 + ||a|| 2 ||c|| 2 



1.6 | Some n-dimensional Geometry 



= (»-»)(^) (a.a)+||a|| 2 ||c|| 2 

= (a.b) 2 + ||a|| 2 ||c|| 2 . 
Now, the quantity ||a|| 2 ||c|| 2 is nonnegative. Hence, 

||a|| 2 ||b|| 2 >(a.b) 2 . 
Taking square roots in this last inequality yields the result desired. ■ 

The geometric motivation for this proof of the Cauchy-Schwarz inequality 
comes from Figure 1 .80. 1 




PROOF Strategic use of the Cauchy-Schwarz inequality yields 
||a + b|| 2 = (a + b)-(a + b) 

= a- a + 2a-b + b- b 

< a-a + 2||a|| ||b|| +b-b by(l) 
= ||a|| 2 + 2||a|| ||b|| + ||b|| 2 
= (l|a|| + ||b||) 2 . 

Thus, the result desired holds by taking square roots, since the quantities on both 
sides of the inequality are nonnegative. ■ 

In two or three dimensions the triangle inequality has the following obvious 
proof from which the inequality gets its name: Since ||a||, ||b||, and ||a + b|| can 
be viewed as the lengths of the sides of a triangle, inequality (2) says nothing 
more than that the sum of the lengths of two sides of a triangle must be at least 
as large as the length of the third side, as demonstrated by Figure 1.81. 

Matrices 

We had a brief glance at matrices and determinants in § 1 .4 in connection with the 
computation of cross products. Now it's time for another look. 

A matrix is defined in § 1 .4 as a rectangular array of numbers. To extend our 
discussion, we need a good notation for matrices and their individual entries. We 
used the uppercase Latin alphabet to denote entire matrices and will continue to 
do so. We shall also adopt the standard convention and use the lowercase Latin 
alphabet and two sets of indices (one set for rows, the other for columns) to 
identify matrix entries. Thus, the general m x n matrix can be written as 



A = 



a\\ an ■•■ din 
C?21 an ■ ■ ■ ain 

- "ml @m2 ' ' ' @mn 



= (shorthand) (fly). 



1 See J. W. Cannon, Amer. Math. Monthly 96 (1989), no. 7, 630-631. 



Chapter 1 | Vectors 



The first index always will represent the row position and the second index, the 
column position. 

Vectors in R" can also be thought of as matrices. We shall have occasion to 
write the vector a = (a\, a%, . . . , a n ) either as a row vector (a 1 x n matrix), 

a = [ a\ ai ■ ■ ■ a n ] , 

or, more typically, as a column vector (an nxl matrix), 

a\ 

a 2 

a = 



1- "71 — I 



We did not use double indices since there is only a single row or column present. 
It will be clear from context (or else indicated explicitly) in which form a vector 
a will be viewed. An m x n matrix A can be thought of as a "vector of vectors" 
in two ways: (1) as m row vectors in R", 

[ an an ■ ■ ■ a\ n ] 
[ <221 a 2 2 ■ ■ ■ a 2n ] 



A 



[ a m \ a m2 
or (2) as n column vectors in R" ! , 



nn 1 





a n 




an 




a\ n 






a 2 \ 








a2 n 






- a m \ - 




- a m 2 _ 




- a mn _ 





A = 



We now define the basic matrix operations. Matrix addition and scalar mul- 
tiplication are really no different from the corresponding operations on vectors 
(and moreover, they satisfy essentially the same properties). 



DEFINITION 6.2 (Matrix Addition) Let A and B be two m x n matrices. 
Then their matrix sum A + B is the m x n matrix obtained by adding cor- 
responding entries. That is, the entry in the ith row and 7 th column of A + B 
is a t j + b t j, where a, 7 and are the ijth entries of A and B, respectively. 



EXAMPLE 2 If 

A = 

then 



However, if B 



and B 



A + B = 



2 
10 



the same dimensions as A. 



, then A + B is not defined, since B does not have 



1.6 | Some n-dimensional Geometry 



Properties of matrix addition. For all m x n matrices A, B, and C we have 

1. A + B = B + A (commutativity); 

2. A + (B + C) = (A + B) + C (associativity); 

3. An m x n matrix O (the zero matrix) with the property that A + O = A 
for all m x n matrices A. 



DEFINITION 6.3 (Scalar Multiplication) If A is an m x n matrix and 
k e R is any scalar, then the product kA of the scalar k and the matrix A is 
obtained by multiplying every entry in A by k. That is, the i/th entry of kA 
is kciij (where ay is the iyth entry of A). 



EXAMPLE 3 IfA = 



1 

4 



, then 3A = 



3 6 c 
12 15 If 



Properties of scalar multiplication. If A and B are any m x n matrices 
and k and / are any scalars, then 

1. (k + l)A = kA + lA (distributivity); 

2. k(A + B) = kA + kB (distributivity); 

3. k(lA) = (kl)A = l(kA). 



We leave it to you to supply proofs of these addition and scalar multiplication 
properties if you wish. 

Just as defining products of vectors needed to be "unexpected" in order to be 
useful, so it is with defining products of matrices. To a degree, matrix multipli- 
cation is a generalization of the dot product of two vectors. 



DEFINITION 6.4 (Matrix Multiplication) Let A be an m x n matrix 
and B an n x p matrix. Then the matrix product A B is the m x p matrix 
whose i/th entry is the dot product of the j'th row of A and the y'th column 
of B (considered as vectors in R"). That is, the ijth entry of 




au ayi ■ ■ ■ a\ n 
[an a i2 ■ ■ ■ a in ] 




-b n ... 

b 2 \ 


~b,jl 
bij 


. . . bi p 

bi p 










.b n \ 


- bnj _ 


Kp. 




is 














a 


i\b\j + a i2 b 2 j H h a 


'n 


n 

b n j = (more compactly) gaby ■ 

k=l 



54 Chapter 1 | Vectors 



EXAMPLE 4 If 

A = 



1 2 3 
4 5 6 



and B = 



0 1 
7 0 
2 4 



then the (2, 1) entry of AB is the dot product of the second row of A and the first 
column of B: 



(2, 1) entry = [ 4 5 6 ] 



= (4)(0) + (5)(7) + (6)(2) = 47. 



The full product A B is the 2x2 matrix 



20 13 
47 28 

On the other hand B A is the 3x3 matrix 



4 5 6 
7 14 21 
18 24 30 



Order matters in matrix multiplication. To multiply two matrices we must 



have 

Number of columns of left matrix = number of rows of right matrix. 

In Example 4, the products AB and BA are matrices of different dimensions; 
hence, they could not possibly be the same. A worse situation occurs when the 
matrix product is defined in one order and not the other. For example, if A is 2 x 3 
and £ is 3 x 3, then A B is defined (and is a 2 x 3 matrix), but B A is not. However, 
even if both products AB and BA are defined and of the same dimensions (as is 
the case if A and B are both n x n, for example), it is in general still true that 

AB + BA. 

Despite this negative news, matrix multiplication does behave well in a number 
of respects, as the following results indicate: 



Properties of matrix multiplication. Suppose A, B, and C are matrices 
of appropriate dimensions (meaning that the expressions that follow are all 
defined) and that k is a scalar. Then 

1. A(BC) = (AB)C; 

2. k(AB) = (kA)B = A(kB); 

3. A(B + C) = AB + AC; 

4. (A + B)C = AC + BC. 



The proofs of these properties involve little more than Definition 6.4, although 
the notation can become somewhat involved as in the proof of property 1 . 



One simple operation on matrices that has no analogue in the real number 
system is the transpose. The transpose of an m x n matrix A is the n x m matrix 



1.6 | Some n-dimensional Geometry 



A T obtained by writing the rows of A as columns. For example, if 

then A T = 



A = 



1 

4 



More abstractly, the ijth entry of A T is a,-,-, the jith entry of A. 

The transpose operation turns row vectors into column vectors and vice versa 
We also have the following results: 

(A T f 



A, 

(A B) 1 = B T A T , 



for any matrix A. (3) 
where A is m x n and B is n x p. (4) 



The transpose will largely function as a notational convenience for us. For 
example, consider a, b £ R" to be column vectors. Then the dot product a • b can 
be written in matrix form as 



■ b = a\b\ + a 2 b 2 + • • • + a„b n = [ a\ 



C'2 



In ] 



b 2 



= a r b. 



EXAMPLE 5 Matrix multiplication is defined the way it is so that, roughly 
speaking, working with vectors or quantities involving several variables can be 
made to look as much as possible like working with a single variable. This idea 
will become clearer throughout the text, but we can provide an important example 
now. A linear function in a single variable is a function of the form f(x) = ax 
where a is a constant. The natural generalization of this to higher dimensions 
is a linear mapping F: R" — > R", F(x) = Ax, where A is a (constant) m x n 
matrix and x £ R". More explicitly, F is a function that takes a vector in R" 
(written as a column vector) and returns a vector in R" ! (also written as a column). 
That is, 



F(x) = Ax = 



an 


an ■ 






X\ 


Cl2l 


a 2 2 ■ 








_ a m \ 


@m2 


&mn _ 







The function F has the properties that F(x + y) = F(x) + F(y) for all x, y e R" 
and F(kx) = k¥(x) for all x e R", i £ R. These properties are also satisfied by 
f(x) = ax, of course. Perhaps more important, however, is the fact that linear 
mappings behave nicely with respect to composition. Suppose F is as just defined 
and G: R" ! — > R p is another linear mapping defined by G(x) = Bx, where B is a 
p x m matrix. Then there is a composite function GoF: R" — > R p defined by 

G o F(x) = G(F(x)) = G(Ax) = fi(Ax) = (BA)x 

by the associativity property of matrix multiplication. Note that BA is defined 
and is a p x n matrix. Hence, we see that the composition of two linear mappings 
is again a linear mapping. Part of the reason we defined matrix multiplication the 
way we did is so that this is the case. ♦ 



EXAMPLE 6 We saw that by interpreting equation (1) in §1.2 inn dimensions, 
we obtain parametric equations of a line in R" . Equation (2) of § 1 .5, the equation 



56 Chapter 1 | Vectors 

for a plane in R 3 through a given point (xo, yo, zo) with given normal vector 
n = Ai + B] + Ck, can also be generalized to n dimensions: 

A x (x x - bi) + A 2 (x 2 — b 2 ) H h A n (x„ - b n ) = 0. 

If we let A = (Ai, A 2 , . . . , A„), b = (b\, &2> • • • > b„) ("constant" vectors), and 
x = (jci, x%, ■ ■ ■ , x„) (a "variable" vector), then the aforementioned equation can 
be rewritten as 

A • (x - b) = 0 

or, considering A, b, and x as n x 1 matrices, as 

A r (x - b) = 0. 

This is the equation for a hyperplane in R" through the point b with normal 
vector A. The points x that satisfy this equation fill out an (n — l)-dimensional 
subset of R". ♦ 

At this point, it is easy to think that matrix arithmetic and the vector geometry 
of R", although elegant, are so abstract and formal as to be of little practical use. 
However, the next example, from the field of economics, 2 shows that this is not 
the case. 

EXAMPLE 7 Suppose that we have n commodities. If the price per unit of the 
ith commodity is pi, then the cost of purchasing x, (> 0) units of commodity 
i is piXj. If p = (pi, .. . , p„) is the price vector of all the commodities and 
x = (jti, . . . , x n ) is the commodity bundle vector, then 

p • x = p x x\ + p 2 x 2 H V p»x„ 

represents the total cost of the commodity bundle. 

Now suppose that we have an exchange economy, so that we may buy and 
sell items. If you have an endowment vector w = (w\ ,...,«>„), where to,- is the 
amount of commodity / that you can sell (trade), then, with prices given by the 
price vector p, you can afford any commodity bundle x where 

p • x < p • w. 

We may rewrite this last equation as 

p • (x - w) < 0. 

In other words, you can afford any commodity bundle x in the budget set 
{x | p • (x — w) < 0}. The equation p • (x — w) = 0 defines a budget hyperplane 
passing through w with normal vector p. ♦ 



Determinants 

We have already defined determinants of 2 x 2 and 3x3 matrices. (See §1.4.) 
Now we define the determinant of any n x n (square) matrix in terms of determi- 
nantsof(n — 1) x (n — 1) matrices. By "iterating the definition," we can calculate 
any determinant. 



2 See D. Saari, "Mathematical complexity of simple economics," Notices of the American Mathematical 
Society 42 (1995), no. 2, 222-230. 



1.6 | Some n-dimensional Geometry 



DEFINITION 6.5 Let A = (ay) be an n x n matrix. The determinant of 
A is the real number given by 

l) 1+2 «i2|A 12 | + --- + (-l) 1+B ai„|A 1 „|, 
1) submatrix of A obtained by deleting the 



\A\={-X) l+l a n \A n 



where Ay is the (n — 1) x (« 
rth row and jth column of A. 



EXAMPLE 8 IfA = 



According to Definition 6.5, 



det 



1 


2 


1 


3 


-2 


1 


0 


5 


4 


2 


-1 


0 


3 


-2 


1 


1 



, then 



+- 




i— 


—i— 




-2 






0 


5 


4 




> 


-1 


0 


3 




> 


1 


1 



2 


0 


5 


4 


-1 


0 


3 


1 


1 



1 


2 


1 


3 " 


-2 


1 


0 


5 


4 


2 


-1 


0 


3 


-2 


1 


1 





1 


0 


5 


(-l) 1+1 (l)det 


2 


-1 


0 




-2 


1 


1 







-2 


0 


5 


+ (- 


l) 1+2 (2)det 


4 


-1 


0 






3 


1 


1 






" -2 


1 


5 


+ (- 


l) 1+3 (l)det 


4 


2 


0 






3 


-2 


1 






" -2 


1 


0 


+ (- 


l) 1+4 (3)det 


4 


2 


-1 






3 


-2 


1 



(l)(l)(-l) + (-l)(2)(37) + (l)(l)(-78) 
+ (-l)(3)(-7) 

-132. ♦ 



The determinant of the submatrix A/j of A is called the ijth minor of A, and 
the quantity (— 1) ,+; |A, 7 | is called the ijth cofactor. Definition 6.5 is known as 
cofactor expansion of the determinant along the first row, since det A is written 
as the sum of the products of each entry of the first row and the corresponding 
cofactor (i.e., the sum of the terms a\j times (— |Ay |). 

It is natural to ask if one can compute determinants by cofactor expansion 
along other rows or columns of A. Happily, the answer is yes (although we shall 
not prove this). 



58 Chapter 1 | Vectors 



Convenient Fact. The determinant of A can be computed by cofactor ex- 


pansion along any row or column. That is, 




\A\ = (-ly+'anl A n | + (-l) i+2 a i2 \A i2 \ + ■ ■ 


■ + (-i) i+n a in \A in \ 


(expansion along the rth row), 




|A| = (-l)!+^ ly .|A ly -| + (-l) 2+J a 2j \A 2j \ + • 


■ + {-X) n+ >a nl \A nj \ 


(expansion along the y'th column). 





EXAMPLE 9 To compute the determinant of 



- 1 


2 


0 


4 


5 _ 


2 


0 


0 


9 


0 


7 


5 


1 


-1 


0 


0 


2 


0 


0 


2 


3 


1 


0 


0 


0 



expansion along the first row involves more calculation than necessary. In partic- 
ular, one would need to calculate four 4x4 determinants on the way to finding 
the desired 5x5 determinant. (To make matters worse, these 4x4 determinants 
would in turn, need to be expanded also.) However, if we expand along the third 
column, we find that 

det A = (- l) 1+3 (0) det A 13 + (- 1) 2+3 (0) det A 23 + (- 1) 3+3 (1) det A 33 

+ (- 1) 4+3 (0) det A 43 + (- 1) 5+3 (0) det A 53 

= det A33 



1 


2 


4 


5 


2 


0 


9 


0 


0 


2 


0 


2 


3 


1 


0 


0 



There are several good ways to evaluate this 4x4 determinant. We'll expand 
about the bottom row: 



12 4 5 

2 0 9 0 
0 2 0 2 

3 10 0 





2 


4 5 




1 


4 5 


(-1) 4+1 (3) 


0 


9 0 


+ (-l) 4+2 (l) 


2 


9 0 




2 


0 2 




0 


0 2 



= (-l)(3)(-54) + (l)(l)(2) 
= 164. 



Of course, not all matrices contain well-distributed zeros as in Example 9, 
so there is by no means always an obvious choice for an expansion that avoids 
much calculation. Indeed one does not compute determinants of large matrices 
by means of cofactor expansion. Instead certain properties of determinants are 
used to make hand computations feasible. Since we shall rarely need to consider 
determinants larger than 3 x 3, we leave such properties and their significance to 
the exercises. (See, in particular, Exercises 26 and 27.) 



1.6 | Exercises 59 



1.6 Exercises 



1 . Rewrite in terms of the standard basis for R" : 

(a) (1,2,3, ...,n) 

(b) (1,0,-1, 1,0,-1,..., 1,0, -1) (Assume that n 
is a multiple of 3.) 

In Exercises 2—4 write the given vectors without recourse to 
standard basis notation. 

2. ei + e2 H he,, 

3. ei - 2e 2 + 3e 3 - 4e 4 H h (-l)" +1 ne„ 

4. ei + e„ 

5. Calculate the following, where a = (1, 3, 5, . . . , 
In - 1) and b = (2, -4, 6, ... , (-l)" +1 2w): 



(a) a + b 
(d) ||a|| 



(b) a - b 
(e) a-b 



(c) -3a 



6. Let n be an even number. Verify the triangle in- 
equality in R" for a = (1, 0, 1, 0, ... , 0) and b = 
(0, 1, 0, 1, ... , 1). 

7. Verify that the Cauchy-Schwarz inequality holds for 
the vectors a = (1 , 2, . . . , n) and b = ( 1 , 1 , . . . , 1). 

8. If a = (1,-1,7, 3, 2) and b = (2, 5, 0, 9, -1), calcu- 
late the projection proj a b. 

9. Show, for all vectors a, b, c e R", that 

|| a — b|| < || a — c|| + ||c — b||. 

10. Prove the Pythagorean theorem. That is, if a, b, and 
c are vectors in R" such that a + b = c and a • b = 0, 
then 

||a|| 2 + ||b|| 2 = ||c|| 2 . 
Why is this called the Pythagorean theorem? 

11. Let a and b be vectors in R". Show that if ||a + b|| = 
|| a — b||, then a and b are orthogonal. 

12. Let a and b be vectors in R". Show that if ||a — b|| > 
|| a + b || , then the angle between a and b is obtuse (i.e., 
more than jt/2). 

13. Describe "geometrically" the set of points in R 5 satis- 
fying the equation 

2( Xl - 1) + 3(x 2 + 2) - 7x 3 + x 4 - 4 - 5(x 5 + 1) = 0. 

14. To make some extra money, you decide to print four 
types of silk-screened T-shirts that you sell at various 
prices. You have an inventory of 20 shirts that you can 
sell for $8 each, 30 shirts that you sell for $10 each, 24 
shirts that you sell for $ 12 each, and 20 shirts that you 
sell for $ 1 5 each. A friend of yours runs a side business 
selling embroidered baseball caps and has an inventory 



of 30 caps that can be sold for $10 each, 16 caps that 
can be sold for $10 each, 20 caps that can be sold for 
$12 each, and 28 caps that can be sold for $15 each. 
You suggest swapping half your inventory of each type 
of T-shirt for half his inventory of each type of baseball 
cap. Is your friend likely to accept your offer? Why or 
why not? 

15. Suppose that you run a grain farm that produces six 
types of grain at prices of $200, $250, $300, $375, 
$450, $500 per ton. 

(a) If x = (xi, . . . , xg) is the commodity bundle vec- 
tor (meaning that x,- is the number of tons of grain 
i to be purchased), express the total cost of the 
commodity bundle as a dot product of two vectors 
inR 6 . 

(b) A customer has a budget of $100,000 to be used 
to purchase your grain. Express the set of possible 
commodity bundle vectors that the customer can 
afford. Also describe the relevant budget hyper- 
plane in R 6 . 

In Exercises 16-19, calculate the indicated matrix quantities 
where 



C 



i : 

-2 ( 

1 -1 

2 0 
0 3 



16. 3A-2S 
18. DB 



, D = 

17. AC 
19. B T D 



20. The n x n identity matrix, denoted I or 7„, is the ma- 
trix whose nth entry is 1 and whose other entries are 
all zero. That is, 



0 0 



(a) Explicitly write out I 2 , h, and I4. 

(b) The reason / is called the identity matrix is that 
it behaves as follows: Let A be any m x n matrix. 
Then 

i. AI„ = A. 

ii. I m A = A. 

Prove these results. (Hint: What are the yth entries 
of the products in (i) and (ii)?) 



Chapter 1 | Vectors 



Evaluate the determinants given in Exercises 21—23. 



22. 



-7 

/ 


A 

u 


1 

— 1 


0 






2 


0 


1 


3 






1 


-3 


0 


2 






0 


5 


1 


-2 






o 
0 


A 


A 
U 


0 




1 c 


1 
1 


A 

u 


0 




7 


0 


1 

— 1 


0 




8 


1 


9 


7 




5 


-1 


0 


8 


11 


0 


2 


1 


9 




7 


0 


0 


4 


-3 




5 


0 


0 


0 


2 




1 


0 


0 


0 


0 




-3 



24. Prove that a matrix that has a row or a column con- 
sisting entirely of zeros has determinant equal to 
zero. 

25. An upper triangular matrix is an n x n matrix whose 
entries below the main diagonal are all zero. (Note: 
The main diagonal is the diagonal going from upper 
left to lower right.) For example, the matrix 



- 1 


2 


-1 


2 " 


0 


3 


4 


3 


0 


0 


5 


6 


_ 0 


0 


0 


7 _ 



is upper triangular. 

(a) Give an analogous definition for a lower triangu- 
lar matrix and also an example of one. 

(b) Use cofactor expansion to show that the determi- 
nant of any n x n upper or lower triangular matrix 
A is the product of the entries on the main diagonal. 
That is, det A = awa^i ■ ■ ■ a„„ . 

26. Some properties of the determinant. Exercises 24 
and 25 show that it is not difficult to compute de- 
terminants of even large matrices, provided that the 
matrices have a nice form. The following operations 
(called elementary row operations) can be used to 
transform annxn matrix into one in upper triangular 
form: 

I. Exchange rows i and j. 
II. Multiply row by a nonzero scalar. 

III. Add a multiple of row i to row j. (Row i remains 

unchanged.) 
For example, one can transform the matrix 

" 0 2 3 " 

1 7 -2 
_ 1 5 9 _ 

into one in upper triangular form in three steps: 



Step 1 . Exchange rows 1 and 2 (this puts a nonzero 
entry in the upper left corner): 



0 


2 


3 " 




1 


7 


-2 


1 


7 


-2 




0 


2 


3 


1 


5 


9 




1 


5 


9 



Step 2. Add —1 times row 1 to row 3 (this eliminates 
the nonzero entries below the entry in the upper left 
corner): 



Step 3 



1 


7 -2 




" 1 


7 


-2 


0 


2 3 




0 


2 


3 


1 


5 9 




0 


-2 


11 




Add row 2 to row 3: 








1 


7 -2 




" 1 


7 


-2 


0 


2 3 




0 


2 


3 


0 


-2 11 




0 


0 


14 



The question is, how do these operations affect the de- 
terminant? 

(a) By means of examples, make a conjecture as to the 
effect of a row operation of type I on the determi- 
nant. (That is, if matrix B results from matrix A by 
performing a single row operation of type I, how are 
det A and det B related?) You need not prove your 
results are correct. 

(b) Repeat part (a) in the case of a row operation of 
type III. 

(c) Prove that if B results from A by multiplying the 
entries in the ;'th row of A by the scalar c (a type II 
operation), then det B = c ■ det A. 

27. Calculate the determinant of the matrix 



" 2 


1 


-2 


7 


8 


1 


0 


1 


-2 


4 


-1 


1 


2 


3 


-5 


0 


2 


3 


1 


7 


_ -3 


2 


-1 


0 


1 



by using row operations to transform A into a matrix 
in upper triangular form and by using the results of 
Exercise 26 to keep track of how the determinant of A 
and the determinant of your final matrix are related. 

28. (a) Is det(A + B) = det A + det 5? Why or why not? 
(b) Calculate 

1 2 7 

3+2 1-1 5+1 
0-2 0 

and 



1 


2 


7 




1 


2 


7 


3 


1 


5 


+ 


2 


-1 


1 


0 


-2 


0 




0 


-2 


0 



and compare your results. 



1.6 | Exercises 61 



1 


3 


2 + 3 


0 


4 


-1+5 


-1 


0 


0-2 



1 


3 


2 




1 


3 


3 


0 


4 


-1 


+ 


0 


4 


5 


-1 


0 


0 




-1 


0 


-2 



(c) Calculate 



and 



and compare your results. 

(d) Conjecture and prove a result about sums of deter- 
minants. (You may wish to construct further exam- 
ples such as those in parts (b) and (c).) 

29. It is a fact that, if A and B are any n x n matrices, then 

det(AB) = (detA)(detfi). 

Use this fact to show that det(A5) = det(BA). (Recall 
that AB / BA, in general.) 

An n x n matrix A is said to be invertible (or nonsingular) if 

there is another n x n matrix B with the property that 

AB = BA = I„, 

where I n denotes the n x n identity matrix. (See Exercise 20.) 
The matrix B is called an inverse to the matrix A. Exercises 
30-38 concern various aspects of matrices and their inverses. 



30. (a) Verify that 



(b) Verify that 



1 0 
1 1 



is an inverse of 



1 0 
-1 1 



1 2 3 

2 5 3 
1 0 8 



is an inverse of 



-40 16 9 

13 -5 -3 

5 -2 -1 

31. Using the definition of an inverse matrix, find an 

2 2 1 

inverse to 0 1 0 

0 0-1 



32. Try to find an inverse matrix to 
What happens? 



2 1 
1 0 
0 -1 



33. Show that if an n x n matrix A is invertible, then A can 
have only one inverse matrix. Thus, we may write A -1 
to denote the unique inverse of a nonsingular matrix 
A. (Hint: Suppose A were to have two inverses B and 
C. Consider B(AC).) 

34. Suppose that A and B are n x « invertible matrices. 
Show that the product matrix A 5 is invertible by ver- 
ifying that its inverse (AS) -1 = B~ l A~ l . 



35. (a) Show that if A is invertible, then det A / 0. (In 

fact, the converse is also true.) 

(b) Show that if A is invertible, then 

det(A" 1 )= — — . 
v ; detA 

36. (a) Show that, if ad — be / 0, then a general 2x2 
a b 



matrix 



1 



c d 



has the matrix 



ad — be 



d 

-c 



ad— be 

c 

ad— be 



as inverse. 

(b) Use this formula to find an inverse of 



ad— be 
a 

ad— be 



2 4 
-1 2 



37. If A is a 3 x 3 matrix and det A / 0, then there is 
a (somewhat complicated) formula for A -1 . In 
particular, 



detA 



|A„| 
-IA12I 
|An| 



-IA21I 
IA22I 
-IA23I 



IA31I 
-IA32I 
IA33I 



where Ay denotes the submatrix of A obtained by 
deleting the ;th row and 7th column (see Defini- 
tion 6.5). Use this formula to find the inverse of 



2 1 1 

0 2 4 

1 0 3 



More generally, if A is any n x n matrix and det A^0, 
then 



detA 



where adj A is the adjoint matrix of A, that is, the 
matrix whose ijth entry is (—\) 1+ -'\Aji\. (Note: The 
formula for the inverse matrix using the adjoint is typi- 
cally more of theoretical than practical interest, as there 
are more efficient computational methods to determine 
the inverse, when it exists.) 

38. Repeat Exercise 37 with the matrix 



2-1 3 
1 2 -2 
3 0 1 



Cross products in R". Although it is not possible to define a 
cross product of two vectors in R" as we did for two vectors 
in R 3 , we can construct a "cross product" of n — 1 vectors 
in R" that behaves analogously to the three-dimensional cross 



62 Chapter 1 [ Vectors 



product. To be specific, if 

ai = (an, a u , a in ), a 2 = (021, 022, ■ ■ • , a 2 „), 

= (fln-u, in-12, ■ ■ ■ , a„-\„) 

are n — 1 vectors in R", we define ai X a2 X • • • X a„_i to be 
the vector in R" given by the symbolic determinant 



ai x a2 x • • • x a„_i 



"21 



a„-i 1 



e 2 
"12 
"22 



a„-i2 



e„ 

«2» 



(Here t\ , . . . , e„ are £/ze standard basis vectors for R" .) Exer- 
cises 39-42 concern this generalized notion of cross product. 

39. Calculate the following cross product in R 4 : 

(1, 2, -1, 3) x (0, 2, -3, 1) x (-5, 1, 6, 0). 

40. Use the results of Exercises 26 and 28 to show that 

(a) ai x • • • x a,- x • • • x a^ x • • • x a„_i 

= — (ai x • • • x a j x • • • x a,- x • • • x a„_i ), 
I <i <n — 1 , 1 < j < n — 1 

(b) ai x • • • x kHi x • • • x a„_i 

= &(ai x • • • x a,- x • • • x a„_i), 
1 < i < n — 1 . 



(c) ai x • • • x (a,- + b) x • • • x a„_i 

= ai x • • • x a,- x • • • x a„_i + 
aix---xbx---x a„_i, 
1 <i <n - l.allbeR". 

(d) Show that if b = (b\ , . . . , b„) is any vector in R" , 
then 

b • (ai x a 2 x • • • x a„_i) 
is given by the determinant 



All 
a B -n 



b„ 



41. Show that the vector b = ai x a2 x • • • x a„_i is 
orthogonal to ai , . . . , a„_i . 

42. Use the generalized notion of cross products to 
find an equation of the (four-dimensional) hyper- 
plane in R 5 through the five points Po(l, 0, 3, 0, 4), 
^(2,-1,0,0,5), P 2 (7, 0, 0, 2, 0), P 3 (2,0,3,0,4), 
and P 4 (l,~ 1,3,0, 4). 



Figure 1 .82 The Cartesian 
coordinate system. 



Location y 



(x,y) 



Location x 

Figure 1 .83 Locating a point P, 
using Cartesian coordinates. 



1 .7 New Coordinate Systems 

We hope that you are comfortable with Cartesian (rectangular) coordinates for R 2 
or R 3 . The Cartesian coordinate system will continue to be of prime importance to 
us, but from time to time, we will find it advantageous to use different coordinate 
systems. In R 2 , polar coordinates are useful for describing figures with circular 
symmetry. In R 3 , there are two particularly valuable coordinate systems besides 
Cartesian coordinates: cylindrical and spherical coordinates. As we shall see, 
cylindrical and spherical coordinates are each a way of adapting polar coordinates 
in the plane for use in three dimensions. 

Cartesian and Polar Coordinates on R 2 

You can understand the Cartesian (or rectangular) coordinates (x , y) of a point 
P in R 2 in the following way: Imagine the entire plane filled with horizontal and 
vertical lines, as in Figure 1.82. Then the point P lies on exactly one vertical line 
and one horizontal line. The x -coordinate of P is where this vertical line intersects 
the x-axis, and the y-coordinate is where the horizontal line intersects the y-axis. 
(See Figure 1.83.) (Of course, we've already assigned coordinates along the axes 
so that the zero point of each axis is at the point of intersection of the axes. We 
also normally mark off the same unit distance on each axis.) Note that, because 
of this geometry, every point in R 2 has a uniquely determined set of Cartesian 
coordinates. 

Polar coordinates are defined by considering different geometric informa- 
tion. Now imagine the plane filled with concentric circles centered at the origin 
and rays emanating from the origin. Then every point except the origin lies on 



1.7 | New Coordinate Systems 63 



Figure 1 .86 Locating the point 
with polar coordinates (r, 9), 
where r < 0. 



0 


r = 6 cos 0 


0 


6 


jr/6 


3V3 


n/4 


3V2 


7T/3 


3 


n/2 


0 


2jt/3 


-3 


3jt/4 


-3V2 


5?r/6 


-3V3 


7T 


-6 


77T/6 


-3V3 


57T/4 


-3V2 


4tt/3 


-3 


3^/2 


0 


57T/3 


3 


7tt/4 


372 





Figure 1 .84 The polar coordinate 
system. 



Figure 1 .85 Locating a point P, 
using polar coordinates. 



exactly one such circle and one such ray. The origin itself is special: No circle 
passes through it, and all the rays begin at it. (See Figure 1 .84.) For points P other 
than the origin, we assign to P the polar coordinates (r, 9), where r is the radius 
of the circle on which P lies and 9 is the angle between the positive x-axis and 
the ray on which P lies. (9 is measured as opening counterclockwise.) The origin 
is an exception: It is assigned the polar coordinates (0, 9), where 9 can be any 
angle. (See Figure 1.85.) As we have described polar coordinates, r > 0 since r 
is the radius of a circle. It also makes good sense to require 0 < 9 < 2jt, for then 
every point in the plane, except the origin, has a uniquely determined pair of polar 
coordinates. Occasionally, however, it is useful not to restrict r to be nonnegative 
and 9 to be between 0 and lit. In such a case, no point of R 2 will be described by 
a unique pair of polar coordinates: If P has polar coordinates (r, 9), then it also 
has (r, 9 + 2nn) and (— r, 9 + (2n + \)tt) as coordinates, where n can be any 
integer. (To locate the point having coordinates (r, 9), where r < 0, construct the 
ray making angle 9 with respect to the positive x-axis, and instead of marching 
| r | units away from the origin along this ray, go | r \ units in the opposite direction, 
as shown in Figure 1.86.) 

EXAMPLE 1 Polar coordinates may already be familiar to you. Nonetheless, 
make sure you understand that the points pictured in Figure 1.87 have the coor- 
dinates indicated. ♦ 



(3V3, kI6) 



(2, 5n/6) 


(2, kI6) 

(5,0) 




(-1, 5tt/6) or 




(1, 1171/6) or 


(3, 3tt/2) < 


(l,-7i/6) 




Figure 1 .87 Figure for 
Example 1. 



Figure 1 .88 The graph of 
r = 6 cos 6 in Example 2. 



EXAMPLE 2 Let's graph the curve given by the polar equation r = 6cos# 
(Figure 1.88). We can begin to get a feeling for the graph by compiling values, as 
in the adjacent tabulation. 



64 Chapter 1 [ Vectors 

Thus, r decreases from 6 to 0 as 9 increases from 0 to ji/2; r decreases from 
0 to —6 (or is not defined if you take r to be nonnegative) as 9 varies from ir/2 
to 7r; r increases from —6 to 0 as 9 varies from n to 3jt/2; and r increases from 
0 to 6 as 9 varies from 3jt/2 to 2ir . To graph the resulting curve, imagine a radar 
screen: As 9 moves counterclockwise from 0 to 2n, the point (r, 9) of the graph is 
traced as the appropriate "blip" on the radar screen. Note that the curve is actually 
traced twice: once as 9 varies from 0 to n and then again as 9 varies from it to 
2jt. Alternatively, the curve is traced just once if we allow only 9 values that yield 
nonnegative r values. The resulting graph appears to be a circle of radius 3 (not 
centered at the origin), and in fact, one can see (as in Example 3) that the graph 
is indeed such a circle. ♦ 



The basic conversions between polar and Cartesian coordinates are provided 


by the following relations: 




Polar to Cartesian: 


J x = r cos9 , 
1 y = r sin 9 ' 


Cartesian to polar: 


\ r 2 = x 2 + y 2 

(tan<9 = y/x - (2) 



Note that the equations in (2) do not uniquely determine r and 9 in terms of x 
and y. This is quite acceptable, really, since we do not always want to insist that 
r be nonnegative and 9 be between 0 and 2n. If we do restrict r and 9, however, 
then they are given in terms of x and y by the following formulas: 

r = y/x 2 + y 2 , 



tan -1 y/x 


if x 


> 


0,y 


> 


0 


tan -1 y/x + 2jt 


ifx 


> 


0,y 


< 


0 


tan -1 y/x + n 


ifx 


< 


0,y 


> 


0 


tt/2 


ifx 




0,y 


> 


0 


3tt/2 


ifx 




0,y 


< 


0 


indeterminate 


ifx 




y = 


0 





The complicated formula for 9 arises because we require 0 < 9 < 2n , while the 
inverse tangent function returns values between — ir/2 and n/2 only. Now you 
see why the equations given in (2) are a better bet! 

EXAMPLE 3 We can use the formulas in (1) and (2) to prove that the curve in 
Example 2 really is a circle. The polar equation r = 6 cos 9 that defines the curve 
requires a little ingenuity to convert to the corresponding Cartesian equation. The 
trick is to multiply both sides of the equation by r. Doing so, we obtain 

r 2 = 6r cosO. 

Now (1) and (2) immediately give 

x 2 + y 2 = 6x. 



1.7 | New Coordinate Systems 65 
We complete the square in x to find that this equation can be rewritten as 

(x - 3) 2 + y 2 = 9, 

which is indeed a circle of radius 3 with center at (3, 0). ♦ 



Cylindrical Coordinates 

Cylindrical coordinates on R 3 are a "naive" way of generalizing polar coordinates 
to three dimensions, in the sense that they are nothing more than polar coordinates 
used in place of the x- and y-coordinates. (The z-coordinate is left unchanged.) 
The geometry is as follows: Except for the z-axis, fill all of space with infinitely 
extended circular cylinders with axes along the z-axis as in Figure 1.89. Then 
any point P in R 3 not lying on the z-axis lies on exactly one such cylinder. 
Hence, to locate such a point, it's enough to give the radius of the cylinder, the 
circumferential angle 9 around the cylinder, and the vertical position z along the 
cylinder. The cylindrical coordinates of P are (r, 9, z), as shown in Figure 1.90. 
Algebraically, the equations in (1) and (2) can be extended to produce the basic 
conversions between Cartesian and cylindrical coordinates. 



The basic conversions between cylindrical 


and Cartesian coordinates 


are 


provided by the following relations: 








x = r cos 6* 




Cylindrical to Cartesian: 


y = r sin 9 ; 

Z = Z 


(3) 




r 2 = x 2 + y 2 




Cartesian to cylindrical: 


tan 9 = y/x . 


(4) 




z = z 





As with polar coordinates, if we make the restrictions r > 0, 0 < 9 < 2n, then 
all points of R 3 except the z-axis have a unique set of cylindrical coordinates. A 
point on the z-axis with Cartesian coordinates (0,0, zo) has cylindrical coordinates 
(0, 9, zo), where 9 can be any angle. 

Cylindrical coordinates are useful for studying objects possessing an axis of 
symmetry. Before exploring a few examples, let's understand the three "constant 
coordinate" surfaces. 

• The r = T(, surface is, of course, just a cylinder of radius ro with axis the 
z-axis. (See Figure 1.91.) 

• The 9 = 9q surface is a vertical plane containing the z-axis (or a half-plane 
with edge the z-axis if we take r > 0 only). (See Figure 1.92.) 

• The z = zo surface is a horizontal plane. (See Figure 1.93.) 



Half-plane only 




EXAMPLE 4 Graph the surface having cylindrical equation r = 6 cos 8. (This 
equation is identical to the one in Example 2.) In particular, z does not appear 
in this equation. What this means is that if the surface is sliced by the horizontal 
plane z = c where c is a constant, we will see the circle shown in Example 2, no 
matter what c is. If we stack these circular sections, then the entire surface is a 
circular cylinder of radius 3 with axis parallel to the z-axis (and through the point 
(3, 0, 0) in cylindrical coordinates). This surface is shown in Figure 1.94. ♦ 

EXAMPLE 5 Graph the surface having equation z = 2r in cylindrical co- 
ordinates. 

Here the variable 6 does not appear in the equation, which means that the 
surface in question will be circularly symmetric about the z-axis. In other words, 
if we slice the surface by any plane of the form 6 = constant (or half-plane, if we 
take r > 0), we see the same curve, namely, a line (respectively, a half-line) of 
slope 2. As we let the constant-0 plane vary, this line generates a cone, as shown 
in Figure 1 .95. The cone consists only of the top half (nappe) when we restrict r 
to be nonnegative. 

The Cartesian equation of this cone is readily determined. Using the formulas 
in (4), we have 

Z = 2r => z 2 =4r 2 <^=> z 2 = 4(x 2 + y 2 ). 

Since z can be positive as well as negative, this last Cartesian equation describes 
the cone with both nappes. If we want the top nappe only, then the equation 
z = 2y 'x 2 + y 2 describes it. Similarly, z = —2^x 2 + y 2 describes the bottom 
nappe. ♦ 

Spherical Coordinates 

Fill all of space with spheres centered at the origin as in Figure 1 .96. Then every 
point P e R 3 , except the origin, lies on a single such sphere. Roughly speaking, 
the spherical coordinates of P are given by specifying the radius p of the sphere 
containing P and the "latitude and longitude" readings of P along this sphere. 
More precisely, the spherical coordinates (p, <p, 9) of P are defined as follows: p 
is the distance from P to the origin; <p is the angle between the positive z-axis and 
the ray through the origin and P; and 6 is the angle between the positive x-axis 
and the ray made by dropping a perpendicular from P to the xy-plane. (See Figure 
1 .97.) The 0 -coordinate is exactly the same as the ^-coordinate used in cylindrical 
coordinates. (Warning: Physicists usually prefer to reverse the roles of <p and 9, 
as do some graphical software packages.) 



1.7 | New Coordinate Systems 67 



z 




Figure 1 .96 The spherical Figure 1 .97 Locating the point 

coordinate system. P, using spherical coordinates. 



It is standard practice to impose the following restrictions on the range of 
values for the individual coordinates: 

P > 0, 0 < <p < n, 0 < 9 < lit. (5) 

With such restrictions, all points of R 3 , except those on the z-axis, have a uniquely 
determined set of spherical coordinates. Points along the z-axis, except for the 
origin, have coordinates of the form (po, 0, 9) or (po, n, 9), where po is a positive 
constant and 9 is arbitrary. The origin has spherical coordinates (0,<p,9), where 
both (p and 6 are arbitrary. 



EXAMPLE 6 Several points and their corresponding spherical coordinates are 
shown in Figure 1.98. ♦ 




Figure 1.98 Figure for Example 6. Figure 1.99 The graph of Figure 1.100 The spherical 

P = Po (> 0). surface <p = (po, shown for 

different values of (po. 

Spherical coordinates are especially useful for describing objects that have 
a center of symmetry. With the restrictions given by the inequalities in (5), the 
constant coordinate surface p = po (po > 0) is, of course, a sphere of radius po, 
as shown in Figure 1.99. The surface given by 9 = 9o is a half-plane just as in the 
cylindrical case. The <p = <po surface is a single-nappe cone if cpo / tt/2 and is 
the xy-plane if cp 0 = tt/2 (and is the positive or negative z-axis if (po = 0 or it). 
(See Figure 1 . 1 00.) If we do not insist that p be nonnegative, then the cones would 
include both nappes. 



68 Chapter 1 [ Vectors 



The basic equations relating spherical coordinates to both cylindrical and 


Cartesian coordinates are as follows. 




Spherical/cylindrical: 








r = p sin cp 


—2 2 i _2 

p = r + z 






9 = 9 


\my = r/z . (6) 






z = p cos cp 


9=9 


Spherical/Cartesian: 






x = psincp cos 9 


p 2 = x 2 + y 2 + z 2 




y = p sin (p sin 9 


tan<p = y /x 2 + y 2 /z- (7) 




z = p coscp 


tan# = y/x 





Figure 1.101 Converting 
spherical to cylindrical coordinates 
when 0 < <p < j. 



Figure 1.102 Converting 
spherical to cylindrical 
coordinates when jt/2 < cp < it . 



Using basic trigonometry, it is not difficult to establish the conversions in (6). 
From the right triangle shown in Figure 1.101, we have 



COS I 



Hence, 
Similarly, 

so that 



r = p cos 



p sin<p. 



sin 



p sin 



(*-')- 

(£-')- 



z 
P 

p cos (p. 



Thus, the formulas in (6) follow when 0 < <p < tt/2. If it/2 < <p < n, then we 
may employ Figure 1.102. So 



p cos 



('-I) 



p sm<p, 



and 



z = -P sin ^ - -) = p sin - <p) = 



p cos<p. 



1.7 | New Coordinate Systems 69 



V 


p = 2a cos <p 


0 


2a 


7T/6 


V3a 


ji/4 


V2a 


tt/3 


a 


JT/2 


0 


2jt/3 


—a 


3jt/4 


— \/2a 


7T 


—2a 



Hence, the relations in (6) hold in general. The equations in (7) follow by substi- 
tution of those in (6) into those of (3) and (4). 

EXAMPLE 7 The cylindrical equation z = 2r in Example 5 converts via (6) 
to the spherical equation 

p cosco = 2p simp. 

Therefore, 



tano5 



26°. 



1 -i 1 
- <=?■ co = tan - 

2 r 2 

Thus, the equation defines a cone (as we just saw). The spherical equation is 
especially simple in that it involves just a single coordinate. ♦ 



EXAMPLE 8 Not all spherical equations are improvements over their cylindri 

y 



cal or Cartesian counterparts. For example, the Cartesian equation 6x = x 2 1 



(whose polar-cylindrical equivalent is r = 6 cos 9) becomes 

6p sin (p cos 9 = p 2 sin 2 05 cos 2 9 + p 2 sin 2 cp sin 2 9 
from (7). Simplifying, 



6p sin (p cos 9 = p sin <p (cos © + sin 9) 
6p sin 09 cos 9 = p 2 sin 2 <p 
6cos# = p sin 05. 



This spherical equation is more complicated than the original Cartesian equation 
in that all three spherical coordinates are involved. Therefore, it is not at all 
obvious that the spherical equation describes a cylinder. ♦ 

EXAMPLE 9 Let's graph the surface with spherical equation p = 2a cosoo, 
where a > 0. As with the graph of the cone with cylindrical equation z = 2r, 
note that the equation is independent of 9. Thus, all sections of this surface made 
by slicing with the half-plane 9 = c must be the same. If we compile values as 
in the adjacent table, then the section of the surface in the half-plane 9 = 0 is as 
shown in Figure 1.103. Since this section must be identical in all other constant-0 
half-planes, we see that this surface appears to be a sphere of radius a tangent to 
the xy-plane, which is shown in Figure 1.104. 




(2a, 0, 0) 

Section of 
p = 2a cos q> 



Figure 1.103 The cross section 
of p = 2a cos 05 in the half-plane 
6 = 0. 




Figure 1 .1 04 The graph of p = 
2a cos<p. 



Chapter 1 | Vectors 



The Cartesian equation of the surface is determined by multiplying both sides 
of the spherical equation by p and using the conversion equations in (7): 

p = 2a cos <p p 2 = lap cos <p 

<^=> x 2 + y 2 + z 2 = 2az 
<^=> x 2 + y 2 + (z - a) 2 = a 2 

by completing the square in z. This last equation can be recognized as that of a 
sphere of radius a with center at (0, 0, a) in Cartesian coordinates. ♦ 

EXAMPLE 10 NASA launches a 10-ft-diameter space probe. Unfortunately, 
a meteor storm pushes the probe off course, and it is partially embedded in 
the surface of Venus, to a depth of one quarter of its diameter. To attempt to 
reprogram the probe's on-board computer to remove it from Venus, it is necessary 
to describe the embedded portion of the probe in spherical coordinates. Let us 
find the description desired, assuming that the surface of Venus is essentially flat 
in relation to the probe and that the origin of our coordinate system is at the center 
of the probe. 














z = -5/2 



Figure 1 .1 07 Coordinate view of 
the cross section of the probe of 
Example 10. 



The situation is illustrated in Figure 1 . 1 05 . The buried part of the probe clearly 
has symmetry about the z-axis. That is, any slice by the half-plane 9 = constant 
looks the same as any other. Thus, 9 can vary between 0 and 2it . A typical slice 
of the probe is shown in Figure 1 . 106. Elementary trigonometry indicates that for 
the angle a in Figure 1.106, 

1 1 
cos a = — = -. 
5 2 

Hence, a = cos -1 \ = n/3. Thus, the spherical angle <p (which opens from the 
positive z-axis) varies from n — tt/3 = 2n/3 to it as it generates the buried part 
of the probe. Finally, note that for a given value of y> between 2n/3 and it, p 
is bounded by the surface of Venus (the plane z = — | in Cartesian coordinates) 
and the spherical surface of the probe (whose equation in spherical coordinates is 
p = 5). See Figure 1.107. From the formulas in (7) the equation z = — f corre- 
sponds to the spherical equation p cos (p = — | or, equivalently, to p = — | sec <p. 
Therefore, the embedded part of the probe may be defined by the set 



5 2tt „ ' 

- seccp < p < 5, — <(p<tt,0<9<2tt 



1.7 | New Coordinate Systems 



Standard Bases for Cylindrical 

and Spherical Coordinates 

In Cartesian coordinates, there are three special unit vectors i, j, and k that point 
in the directions of increasing x-, y-, and z-coordinate, respectively. We find 
corresponding sets of vectors for cylindrical and spherical coordinates. That is, 
in each set of coordinates, we seek mutually orthogonal unit vectors that point in 
the directions of increasing coordinate values. 

In cylindrical coordinates, the situation is as shown in Figure 1.108. The 
vectors e, , e e , and e-, which form the standard basis for cylindrical coordinates, 
are unit vectors that each point in the direction in which only the coordinate 
indicated by the subscript increases. There is an important difference between 
the standard basis vectors in Cartesian and cylindrical coordinates. In the former 
case, i, j, and k do not vary from point to point. However, the vectors e, and eg 
do change as we move from point to point. 

Now we give expressions for e r , eg, and e z . Since the cylindrical z-coordinate 
is the same as the Cartesian z-coordinate, we must have e z = k. The vector 
e r must point radially outward from the z-axis with no k-component. At a 
point (x, y, z) 6 R 3 (Cartesian coordinates), the vector x'\ + y\ has this prop- 
erty. Normalizing it to obtain a unit vector (see Proposition 3.4 of §1.3), we 
obtain 

= xi + y\ 

v 7 * 2 + y 2 ' 

With e r and e- in hand it's now a simple matter to define eg, since it must be 
perpendicular to both e,- and e z . We take 

-yi + xj 

eg = e ; xe r = 

V* 2 + y 2 

(The reason for this choice of cross product, as opposed to e, x e z , is so that eg 
points in the direction of increasing 0.) To summarize, and using the cylindrical 
to Cartesian conversions given in (3), 



e, 


xi + yj 

= — = cos# i + sm6 j; 

V* 2 + y 2 




eg 


-yi + *j . a . . a . 

= — = — sin0 1 + cos0 j; 

V* 2 + v 2 


(8) 




= k. 





In spherical coordinates, the situation is shown in Figure 1 . 109. In particular, 
there are three unit vectors e p , , and eg that form the standard basis for spherical 
coordinates. These vectors all change direction as we move from point to point. 

We give expressions for e p , , and e e . Since the ^-coordinates in both spher- 
ical and cylindrical coordinates mean the same thing, eg in spherical coordinates 
is given by the value of eg in (8). At a point (x,y, z), the vector e p should point 



72 Chapter 1 [ Vectors 

from the origin directly to (x, y, z). Thus, e p may be obtained by normalizing 
■*i + yi + zk. Finally, e p is nothing more than eg x e^. If we explicitly perform 
the calculations just described and make use of the conversion formulas in (7), 
the following are obtained: 





xi + yj + zk 

= — = sm<p cos 6 

Vx 2 + y 2 + z 2 


i + sin <p sin 0 j + cos <p k; 




xzi + yzj - (x 2 + y 2 )k 

v 7 * 2 + y 2 Vx 2 + y 2 + ^ 2 


(9) 

sin <p k; 




= cos <p cos 0 i + cos cp sin 0 j - 




— yi + xj 
= — = — sm 6 1 + cos 6 

V* 2 + 





Although the results of (8) and (9) will not be used frequently, they will prove 
helpful on occasion. 

Hyperspherical Coordinates (optional) 

There is a way to provide a set of coordinates for R" that generalizes spherical 
coordinates on R 3 . For n > 3, the hyperspherical coordinates of a point P e R" 
are (p, q>\, q>%, . . . , <p n -i) and are defined by their relations with the Cartesian 
coordinates (xi, x%, . . . , x n ) of P as 

' xi = p sin sin ip2 ■ ■ ■ sin <p„_2 cos <p„_ i 

x 2 = p sin^i sin ip 2 ■ ■ ■ sin<p„_ 2 sin<p n _i 

x 3 = p sinip! sin • • • sin<p,,_ 3 cos<p„_2 

(10) 

X4 = p sm ipi sm q>2 - - ■ sm <p„_4 cos <p„_3 



_x n = p COS^i 
To be more explicit, in equation (10) above we take 

X* = p sin sin cp 2 - ■ ■ sin cos ^„_jt+i for k = 3, . . . , n. 
Note that when n = 3, the relations in (10) become 

xi = p sin^i cos i^2 
• X2 = p sin^i sini£>2 ■ 
*3 = pcos^i 

These relations are the same as those given in (7), so hyperspherical coordinates 
are indeed the same as spherical coordinates when n = 3. 

In analogy with (5), it is standard practice to impose the following restrictions 
on the range of values for the coordinates: 

P > 0, 0 < (pi < 7T for k = 1, . . . , n — 2, 0 < q> n -\ < lit. (1 1) 



1.7 | Exercises 73 



Then, with these restrictions, we can convert from hypersphencal coordinates to 
Cartesian coordinates by means of the following formulas: 

' p 2 = x \ + x\ H hi„ 2 

tanpi = ^2 H hx2_,/x„ 

tan ^2 = J A H h x*_ 2 /Xn-i n 9 , 



tan<p„_ 2 = J + *f/*3 
tan^„_! = x 2 /*i 

Hyperspherical coordinates get their name from the fact that the (n — 1)- 
dimensional hypersurface in R" defined by the equation p = p 0 , where po is a 
positive constant, consists of points on the hypersphere of radius po defined in 
Cartesian coordinates by the equation 

x 2 +x 2 + ---+x 2 = p 2 . 



1.7 Exercises 



In Exercises 1—3, find the Cartesian coordinates of the points 
whose polar coordinates are given. 

1. (V2,jr/4) 

2. (V3, 5jt/6) 

3. (3, 0) 

In Exercises 4-6, give a set of polar coordinates for the point 
whose Cartesian coordinates are given. 

4. (273,2) 

5. (-2,2) 

6. (-1,-2) 

In Exercises 7-9, find the Cartesian coordinates of the points 
whose cylindrical coordinates are given. 

7. (2,2,2) 

8. (n,n/2, 1) 

9. (1,2* /3, -2) 

In Exercises 10-13, find the rectangular coordinates of the 
points whose spherical coordinates are given. 

10. (4, ;r/2,7r/3) 

11. (3, ;r/3,7r/2) 

12. (1,3tt/4, 2jt/3) 

13. (2,;r,7r/4) 



In Exercises 14-16, find a set of cylindrical coordinates of the 
point whose Cartesian coordinates are given. 

14. (-1,0,2) 

15. (-1, V3, 13) 

16. (5,6,3) 

In Exercises 1 7 and 18, find a set of spherical coordinates of 
the point whose Cartesian coordinates are given. 

17. (1, -1, v/6) 

18. (0, V3, 1) 

19. This problem concerns the surface described by the 
equation (r — 2) 2 + z 2 = 1 in cylindrical coordinates. 
(Assume r > 0.) 

(a) Sketch the intersection of this surface with the half- 
plane 8 = jt/2. 

(b) Sketch the entire surface. 

20. (a) Graph the curve in R 2 having polar equation r = 

2a sin 9, where a is a positive constant. 

(b) Graph the surface in R 3 having spherical equation 
p = 2a sin <p. 

21. Graph the surface whose spherical equation is p = 
1 — cosip. 

22. Graph the surface whose spherical equation is p = 
1 — sinip. 

In Exercises 23-25, translate the following equations from 
the given coordinate system (i.e., Cartesian, cylindrical, or 



74 Chapter 1 | Vectors 



spherical) into equations in each of the other two systems. In 
addition, identify the surfaces so described by providing ap- 
propriate sketches. 

23. p sin<p sin# = 2 

24. z 2 = 2x 2 + 2y 2 

25. r = 0 

In Exercises 26-29, sketch the solid whose cylindrical coordi- 
nates (r, 9, z) satisfy the given inequalities. 

26. 0<r<3, 0<6»<tt/2, -1 < z < 2 

27. r < z < 5, 0 < 9 < n 

28. 2r < z < 5 - 3r 

29. r 2 - 1 < z < 5 - r 2 

In Exercises 30-35, sketch the solid whose spherical coordi- 
nates (p, <p, 9) satisfy the given inequalities. 

30. 1 < p < 2 

31. 0 < p < 1, 0 < (p < tt/2 

32. 0 < p < 1, 0 < 6» < jr/2 

33. 0 < p < jt/4, 0 < p < 2 

34. 0 < p < 2/coscp, 0 < tp < tt/4 

35. 2 cos (p < p <3 

36. (a) Which points P in R 2 have the same rectangular 

and polar coordinates? 

(b) Which points P in R 3 have the same rectangular 
and cylindrical coordinates? 

(c) Which points P in R 3 have the same rectangular 
and spherical coordinates? 

37. (a) How are the graphs of the polar equations r = f(9) 

andr = -f{9) related? 

(b) How are the graphs of the spherical equations 
p = f((p, 9) and p = -f(q>, 0) related? 

(c) Repeat part (a) for the graphs of r = f{9) and 
r = 3/(0). 

(d) Repeat part (b) for the graphs of p = f((p, 9) and 
p = 3f(<p,0). 

38. Suppose that a surface has an equation in cylindrical 
coordinates of the form z = f(r). Explain why it must 
be a surface of revolution. 

39. (a) Verify that the basis vectors e r , eg, and e z for 

cylindrical coordinates are mutually perpendicu- 
lar unit vectors. 

(b) Verify that the basis vectors e p , , and eg for spher- 
ical coordinates are mutually perpendicular unit 
vectors. 



40. Use the formulas in (8) to express i, j, k in terms of e r , 
eg , and e z . 

41 . Use the formulas in (9) to express i, j, k in terms of e p , 
e v , and eg. 

42. Consider the solid in R 3 shown in Figure 1.110. 

(a) Describe the solid, using spherical coordinates. 

(b) Describe the solid, using cylindrical coordinates. 

A portion of the 
z sphere of radius 3 
(centered at origin) 




y 



x 



Figure 1.110 The ice-cream- 
cone-like solid in R 3 in Exercise 42. 

In Exercises 43-47, you will use the equations in (10) to es- 
tablish those in (12). 

43. Show that tan <p n -\ = x%lx\. 

44. (a) Calculate x\ + x\ in terms of the hyperspherical 

coordinates p, (pi, . . . , (p n -%. 

(b) Assuming the inequalities in (1 1), use part (a) to 
show that tamp„_2 = ^jx\ + x 2 /x^. 

45. (a) Calculate x\ + x\ + x 2 in terms of the hyperspher- 

ical coordinates p, <p\, . .. , <p„-z. 

(b) Assuming the inequalities in (1 1), use part (a) to 
show that tan ip,,- 3 = J 1 x\ + x\ + x 2 /x4. 

46. (a) For k — 2, . . . , n — 1, show that x\ + x\ + ■ ■ ■ + 

x\ = p 2 sin 2 <p\ • • ■ sin 2 (p„-k- (Note: This is best 
accomplished by means of mathematical induc- 
tion.) 

(b) Assuming the inequalities in (11), use part (a) 
to show that, for k = 2 n — 1, tan^„_j. = 

jx 2 + ---+xt/x k+l . 

47. Show that x\ + x\ H \- x 2 = p 2 . 



Miscellaneous Exercises for Chapter 1 



True/False Exercises for Chapter 1 



1. If a = (1, 7, -9) and b = (1, -9, 7), then a = b. 

2. If a and b are two vectors in R 3 and k and / are real 
numbers, then (k — Z)(a + b) = ka — /a + kb — lb. 

3. The displacement vector from Pi (1,0, — 1) to 
P 2 (5,3,2) is (-4,-3,-3). 

4. Force and acceleration are vector quantities. 

5. Velocity and speed are vector quantities. 

6. Displacement and distance are scalar quantities. 

7. If a particle is at the point (2, —1) in the plane and 
moves from that point with velocity vector v = (1, 3), 
then after 2 units of time have passed, the particle will 
be at the point (5, 1). 

8. The vector (2,3, —2) is the same as 2i + 3j — 2k. 

9. A set of parametric equations for the line through 
(1, -2, 0) that is parallel to (-2, 4, 7) is x = 1 - 2t, 

y = 4t-2,z = l. 

10. A set of parametric equations for the line through 
(1, 2, 3) and (4, 3, 2) is x = 4 - 3f, y = 3 - t, z = 
t + 2. 

1 1 . The line with parametric equations x = 2 — 
3t, y = t+l, z = 2t — 3 has symmetric form 
x+2 _ _ l _z-3 

-3 ~ y ~ ~ 2 

12. The two sets of parametric equations x = 3t — 1, 
y = 2 - t, z = 2t + 5 and x = 2 - 6t, y = 2t + 1, 
z = 1 — At both represent the same line. 

13. The parametric equations x = 2sinf, y = 2cos?, 
where 0 < t < n, describe a circle of radius 2. 

1 4. The dot product of two unit vectors is 1 . 

15. For any vector a in R" and scalar k, we have ||£a|| = 
fc||a||. 

16. If a, u e R" and ||u|| = 1, then proj u a = (a • u)u. 

17. For any vectors a, b, c in R 3 , we have a x (b x c) = 
(a x b) x c. 



1 8. The volume of a parallelepiped determined by the vec- 
tors a, b, c e R 3 is |(a x c) • b|. 

19. ||a||b — ||b||a is a vector. 

20. (a x b) • c — (a x c) • b is a scalar. 

21. The plane containing the points (1, 2, 1), (3, —1, 0), 
and (1, 0, 2) has equation 5a: + 2y + 4z = 13. 

22. The plane containing the points (1, 2, 1), (3, —1, 0), 
and (1,0,2) is given by the parametric equations 

x = 2s, y = —3s — 2t, z = t — s. 

23. If A is a 5 x 7 matrix and B is a 7 x 7 matrix, then BA 
is a 7 x 5 matrix. 



24. If A 



1 2 0 

-10 2 

5 9 2 

0 8 0 



then 







-1 


2 




1 




1 


0 


3 


detA = 2 


5 


2 




0 


+ 9 


-1 


2 


1 




0 


0 




-6 




0 


0 


-6 






1 


0 


3 








-8 




1 


2 


1 












5 


2 


0 









25. If A is an n x n matrix, then det (2A) = 2 det A. 

26. The surface having equation r = 4 sin 9 in cylindrical 
coordinates is a cylinder of radius 2. 

27. The surface having equation p = 4 cos 6 in spherical 
coordinates is a sphere of radius 2. 

28. The surface having equation p cos 6 sin cp = 3 in spher- 
ical coordinates is a plane. 

29. The surface having equation p = 3 in spherical coor- 
dinates is the same as the surface whose equation in 
cylindrical coordinates is r 2 + z 2 = 9. 

30. The surface whose equation in cylindrical coordinates 
is z = 2r is the same as the surface whose equation in 
spherical coordinates is <p = jt/6. 



Miscellaneous Exercises for Chapter 1 



1 . If P\ , Pi, . . . , P n are the vertices of a regular polygon 
having n sides and if O is the center of the polygon, 

show that Y^=\ OP, = 0. The case n = 5 is shown in 



Figure 1.111. (Hint: Don't try using coordinates. Use 
instead sketches, geometry, and perhaps translations or 
rotations.) 



Chapter 1 | Vectors 




p 4 P, 
Figure 1.111 The case n = 5. 

2. Find parametric equations for the line through the point 
(1, 0, —2) that is parallel to the line x = 3t + 1, y = 
5 - 7r, z = ; + 12. 

3. Find parametric equations for the line through the 
point (1, 0, —2) that intersects the line x = 3t + 1, 
y = 5 — It , z = t + 12 orthogonally. (Hint: Let Xq = 
3% + 1, yo = 5 — ltd, zo = to + 12 be the point where 
the desired line intersects the given line.) 

4. Given two points Po(ai, cti, ai) and P\(b\, £>2, fo), we 
have seen in equations (3) and (4) of §1.2 how to 
parametrize the line through Pq and Pi as r(r) = 
O Pq + t Pq Pi , where t can be any real number. (Recall 
that r = OP, the position vector of an arbitrary point 
P on the line.) 

(a) For what value of / does r(f) = OPb? For what 
value of t does r(r) = OP\l 

(b) Explain how to parametrize the line segment join- 
ing Pq and Pi. (See Figure 1.112.) 



z 




y 



Figure 1.112 The segment joining Po 
and Pi is a portion of the line containing 
Po and P|. (See Exercise 4.) 

(c) Give a set of parametric equations for the line seg- 
ment joining the points (0, 1, 3) and (2, 5, —7). 

5. Recall that the perpendicular bisector of a line seg- 
ment in R 2 is the line through the midpoint of the seg- 
ment that is orthogonal to the segment. 



(a) Give a set of parametric equations for the perpen- 
dicular bisector of the segment joining the points 
Pi(-l,3)andP 2 (5, -7). 

(b) Given general points Pi(ai,fl2) and P2(b\,b2), 
provide a set of parametric equations for the per- 
pendicular bisector of the segment joining them. 

6. If we want to consider a perpendicular bisector of a 
line segment in R 3 , we will find that the bisector must 
be a plane. 

(a) Give an (implicit) equation for the plane that serves 
as the perpendicular bisector of the segment join- 
ing the points P(6, 3, -2) and P 2 (-4, 1, 0). 

(b) Given general points P\{a\, 02, 03) and 
Pi{b\, bj, ^3), provide an equation for the plane 
that serves as the perpendicular bisector of the 
segment joining them. 

7. Generalizing Exercises 5 and 6, we may define the per- 
pendicular bisector of a line segment in R" to be the 
hyperplane through the midpoint of the segment that 
is orthogonal to the segment. 

(a) Give an equation for the hyperplane in R 5 that 
serves as the perpendicular bisector of the seg- 
ment joining the points Pi(l, 6, 0, 3, — 2) and 
P 2 (-3,-2,4, 1,0). 

(b) Given arbitrary points P\(a\ , . . . , a„) and 
Pzibi, , . . , b n ) in R", provide an equation for 
the hyperplane that serves as the perpendicular 
bisector of the segment joining them. 

8. If a and b are unit vectors in R 3 , show that 

|| a x b|| 2 + (a-b) 2 = 1. 

9. (a) If a • b = a • c, does it follow that b = c? Explain 

your answer. 

(b) If a x b = a x c, does it follow that b = c? 
Explain. 

10. Show that the two lines 

h : x = t-3, y=\-2t, z = 2t + 5 
h : x = 4 - It, y = At + 3, z = 6 - At 

are parallel, and find an equation for the plane that 
contains them. 

1 1 . Consider the two planes x + y = 1 and y + z = 1 . 
These planes intersect in a straight line. 

( a) Find the ( acute) angle of intersection between these 
planes. 

(b) Give a set of parametric equations for the line of 
intersection. 

1 2. Which of the following lines whose parametric equa- 
tions are given below are parallel? Are any the same? 
(a) x = At + 6, y = 2 - It, z = 8* + 1 



Miscellaneous Exercises for Chapter 1 



(b) x = 3 - 6t , y = 3f , z = 4 - 9t 

(c) x = 2 - 2t, y = t + 4, z = -4r - 7 

(d) jc = 2f + 4, y = 1 - f , z = 3? - 2 

13. Determine which of the planes whose equations are 
given below are parallel and which are perpendicular. 
Are any of the planes the same? 

(a) 2x + 3y - z = 3 

(b) -6x + 4y - 2z + 2 = 0 

(c) x + y - z = 2 

(d) 10* + I5y - 5z = 1 

(e) 3x - 2y + z = 1 

14. (a) What is the angle between the diagonal of a cube 

and one of the edges it meets? (Hint: Locate the 
cube in space in a convenient way.) 

(b) Find the angle between the diagonal of a cube and 
the diagonal of one of its faces. 

15. Mark each of the following statements with a 1 if you 
agree, — 1 if you disagree: 

(1) Red is my favorite color. 

(2) I consider myself to be a good athlete. 

(3) I like cats more than dogs. 

(4) I enjoy spicy foods. 

(5) Mathematics is my favorite subject. 

Your responses to the preceding "questionnaire" may 
be considered to form a vector in R 5 . Suppose that you 
and a friend calculate your respective "response vec- 
tors" for the questionnaire. Explain the significance of 
the dot product of your two vectors. 

16. The median of a triangle is the line segment that joins 
a vertex of a triangle to the midpoint of the opposite 
side. The purpose of this problem is to use vectors to 
show that the medians of a triangle all meet at a point. 

(a) Using Figure 1.113, write the vectors B M\ and C Mi 
in terms of AB and AC. 




(b) Let P be the point of intersection of BM\ and CMi. 
Write BP and CP in terms of AB and AC. 

(c) Use the fact that CB = CP + Jb = CA + AB to 
show that P must lie two-thirds of the way from B 
to Mi and two-thirds of the way from C to Mi. 

(d) Now use part (c) to show why all three medians must 
meet at P . 

17. Suppose that the four vectors a, b, c, and d in R 3 are 
coplanar (i.e., that they all lie in the same plane). Show 
that then (a x b) x (c x d) = 0. 

18. Show that the area of the triangle, two of whose 
sides are determined by the vectors a and b (see 
Figure 1.114), is given by the formula 

Area = ^7||a|| 2 ||b|| 2 - (a • b) 2 . 




Figure 1.1 14 The triangle in Exercise 18. 

19. Let A(l,3,-1), 5(4,-1,3), C(2, 5,2), and 
D(5, 1, 6) be the vertices of a parallelogram. 

(a) Find the area of the parallelogram. 

(b) Find the area of the projection of the parallelogram 
in the xy -plane. 

20. (a) For the line / in R 2 given by the equation ax + 

by = d, find a vector v that is parallel to /. 

(b) Find a vector n that is normal to / and has first 
component equal to a. 

(c) If Po(-*o, Vo) is any point in R 2 , use vectors to de- 
rive the following formula for the distance from 
P 0 to /: 

\axo + bye — d\ 

Distance from Pq to / = 



Figure 1.113 Two of the three medians 
of a triangle in Exercise 16. 



Va 2 + b 2 

To do this, you'll find it helpful to use Figure 1.115, 
where Pi(xu yi) is any point on I. 

(d) Find the distance between the point (3, 5) and the 
line 8jc — 5y = 2. 

21. (a) If PoC^o, yo, Zo) is any point in R 3 , use vectors 
to derive the following formula for the distance 
from Po to the plane n having equation Ax + 
By + Cz = D: 

\Ax 0 + By 0 + Czo- D\ 



Distance from Pq to fl 



VA 2 + B 2 + C 2 



78 Chapter 1 [ Vectors 



y 





Pa 






\ n / 






I: ax + by = 











Figure 1.115 Geometric construction for 
Exercise 20. 

Figure 1.116 should help. (P\{x\, yi, Zi) is any 



point in n.) 




Distance 






~-*P^J 




Yl:Ax + By + 






X 





Figure 1.116 Geometric construction for 
Exercise 2 1 . 

(b) Find the distance between the point (1,5, —3) and 
the plane x — 2y + 2z + 12 = 0. 

22. (a) Let P be a point in space that is not contained in the 

plane n that passes through the three noncollinear 
points A , B , and C . Show that the distance between 
P and n is given by the expression 

|p-(bxc)| 

lib x o || 

where p = AP, b = A§, and c = AC. 

(b) Use the result of part (a) to find the distance 
between (1, 0, —1) and the plane containing the 
points (1, 2, 3), (2, -3, 1), and (2,-1, 0). 

23. Let A, B, C, and D denote four distinct points in R 3 . 

(a) Show that A, B, and C are collinear if and only if 
Xi x A~t = 0. 

(b) Show that A, B,C, and D are coplanar if and only 
if (AT? x At)-ci> = Q. 

24. Let x = OP, the position vector of a point P in R 3 . 
Consider the equation 

x-k 1 

II || ~ V2' 



Describe the configuration of points P that satisfy the 
equation. 

25. Let a and b be two fixed, nonzero vectors in R 3 , and let 
c be a fixed constant. Explain how the pair of equations, 

a • x = c 
a x x = b , 

completely determines the vector x e R 3 . 

26. (a) Give examples of vectors a, b, c in R 3 that show 

that, in general, it is not true that a x (b x c) = 
(a x b) x c. (That is, the cross product is not as- 
sociative.) 

(b) Use the Jacobi identity (see Exercise 30 of §1.4) 
to show that, for any vectors a, b, c in R 3 , 

a x (b x c) = (a x b) x c 

if and only if 

(c x a) x b = 0. 

27. (a) Given an arbitrary (i.e., not necessarily regular) 

tetrahedron, associate to each of its four triangular 
faces a vector outwardly normal to that face with 
length equal to the area of that face. (See Fig- 
ure 1.117.) Show that the sum of these four vec- 
tors is zero. (Hint: Describe Vi , . . . , V4 in terms of 
some of the vectors that run along the edges of the 
tetrahedron.) 




v 3 



Figure 1.117 The tetrahedron of part 
(a) of Exercise 27. 

(b) Recall that a polyhedron is a closed surface in 
R 3 consisting of a finite number of planar faces. 
Suppose you are given the two tetrahedra shown 
in Figure 1.118 and that face ABC of one is con- 
gruent to face A' B'C' of the other. If you glue the 
tetrahedra together along these congruent faces, 
then the outer faces give you a six-faced polyhe- 
dron. Associate to each face of this polyhedron an 
outward-pointing normal vector with length equal 
to the area of that face. Show that the sum of these 
six vectors is zero. 



Miscellaneous Exercises for Chapter 1 



C 



A 




B B 

Figure 1.118 In Exercise 27(b), glue the two tetrahedra shown along congruent faces. 



28. 



(c) Outline a proof of the following: Given an n-faced 
polyhedron, associate to each face an outward- 
pointing normal vector with length equal to the 
area of that face. Show that the sum of these n 
vectors is zero. 

Consider a right tetrahedron, that is, a tetrahedron 
that has a vertex R whose three adjacent faces are pair- 
wise perpendicular. (See Figure 1.119.) Use the result 
of Exercise 27 to show the following three-dimensional 
analogue of the Pythagorean theorem: If a, b, and 
c denote the areas of the three faces adjacent to R 
and d denotes the area of the face opposite R, then 
d 2 . 




Figure 1.119 The right tetrahedron 
of Exercise 28. The three faces 
containing the vertex R are pairwise 
perpendicular. 

29. (a) Use vectors to prove that the sum of the squares 
of the lengths of the diagonals of a parallelogram 
equals the sum of the squares of the lengths of the 
four sides. 

(b) Give an algebraic generalization of part (a) for R" . 



30. Show that for any real numbers a\ 
we have 

n 2 



■ ■ , a n ,bi, 

n 



31 . To raise a square (n x n) matrix A to a positive integer 
power n, one calculates A" as A ■ A ■ ■ ■ A (n times), 
(a) Calculate successive powers A, A 2 , A 3 , A 4 of the 
1 1 

matrix A = q \ 



(b) Conjecture the general form of A" for the matrix 
A of part (a), where n is any positive integer. 

(c) Prove your conjecture in part (b) using mathemat- 
ical induction. 



32. A square matrix A is called nilpotent if A" 
some positive power n, 

0 1 1 



0 for 



(a) Show that A 



is nilpotent. 



0 0 0 
0 0 0 

(b) Use a calculator or computer to show that A = 
0 0 0 0 0 



is nilpotent. 



^ 33. 



The n x n matrix H n whose ijth entry is l/(i + j — 1) 
is called the Hilbert matrix of order n . 

(a) Write out H2, H], H^, H5, and H^, Use a com- 
puter to calculate their determinants exactly. What 
seems to happen to det H n as n gets larger? 

(b) Now calculate #10 and det H\q. If you use exact 
arithmetic, you should find that det H\q / 0 and 
hence that //10 is invertible. (See Exercises 30-38 
of § 1 .6 for more about invertible matrices.) 

(c) Now give a numerical approximation A for H\q. 
Calculate the inverse matrix B of this approxima- 
tion, if your computer allows. Then calculate AB 
and B A . Do you obtain the 10x10 identity matrix 
ha in both cases? 

(d) Explain what parts (b) and (c) suggest about the 
difficulties in using numerical approximations in 
matrix arithmetic. 



As a child, you may have played with a popular toy called a 
Spirograph®. With it one could draw some appealing geomet- 
ric figures. The Spirograph consists of a small toothed disk with 
several holes in it and a larger ring with teeth on both inside 
and outside as shown in Figure 1.120. You can draw pictures by 
meshing the small disk with either the inside or outside circles 
of the ring and then poking a pen through one of the holes of 
the disk while turning the disk. (The large ring is held fixed.) 



80 Chapter 1 [ Vectors 




An idealized version of the Spirograph can be obtained 
by taking a large circle (of radius a) and letting a small circle 
(of radius b) roll either inside or outside it without slipping. 
A "Spirograph " pattern is produced by tracking a particu- 
lar point lying anywhere on (or inside) the small circle. Exer- 
cises 34-37 concern this set-up. 

34. Suppose that the small circle rolls inside the larger 
circle and that the point P we follow lies on the circum- 
ference of the small circle. If the initial configuration 
is such that P is at (a, 0), find parametric equations 
for the curve traced by P, using angle t from the posi- 
tive .r-axis to the center B of the moving circle. (This 
configuration is shown in Figure 1.121.) The result- 
ing curve is called a hypocycloid. Two examples are 
shown in Figure 1.122. 



y 









a 


(b. y\ 

fp ^\ 
At ( , b\ 













Figure 1.121 The coordinate 
configuration for finding parametric 
equations for a hypocycloid. 



35. Now suppose that the small circle rolls on the outside 
of the larger circle. Derive a set of parametric equa- 
tions for the resulting curve in this case. Such a curve 
is called an epicycloid, shown in Figure 1.123. 

36. (a) A cusp (or corner) occurs on either the hypocy- 

cloid or epicycloid every time the point P on the 
small circle touches the large circle. Equivalently, 



y 




Figure 1.123 An epicycloid with 
a = 4,b= 1. 



this happens whenever the smaller circle rolls 
through 2tt. Assuming that a/b is rational, how 
many cusps does a hypocycloid or epicycloid 
have? (Your answer should involve a and b in some 
way.) 

(b) Describe in words and pictures what happens when 
a/b is not rational. 

37. Consider the original Spirograph set-up again. If we 
now mark a point P at a distance c from the center 
of the smaller circle, then the curve traced by P is 
called a hypotrochoid (if the smaller circle rolls on 
the inside of the larger circle) or an epitrochoid (if 
the smaller circle rolls on the outside). Note that we 
must have b < a, but we can have c either larger or 
smaller than b. (If c < b, we get a "true" Spirograph 
pattern in the sense that the point P will be on the 
inside of the smaller circle. The situation when c > b 
is like having P mounted on the end of an elongated 
spoke on the smaller circle.) Give a set of parametric 
equations for the curves that result in this way. (See 
Figure 1.124.) 

Exercises 38-43 are made feasible through the use of appropri- 
ate software for graphing in polar, cylindrical, and spherical 



Miscellaneous Exercises for Chapter 1 



coordinates. (Note: When using software for graphing in spher- 
ical coordinates, be sure to check the definitions that are used 
for the angles <p and 8.) 




x 



Figure 1.124 The configuration for finding 
parametric equations for epitrochoids. 

38. (a) Graph the curve in R 2 whose polar equation is 

r = cos 28. 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = cos 28. 

(c) Graph the surface in R 3 whose spherical equation 
is p = cos 2<p. 

(d) Graph the surface in R 3 whose spherical equation 
is p = cos 28. 

39. (a) Graph the curve in R 2 whose polar equation is 

r = sin2#. 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = sin 2$. 

(c) Graph the surface in R 3 whose spherical equation 
is p = sin2<p. 

(d) Graph the surface in R 3 whose spherical equation 
is p = sin 28. Compare the results of this exercise 
with those of Exercise 38. 

40. (a) Graph the curve in R 2 whose polar equation is 

r = cos 38. 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = cos 38. 

(c) Graph the surface in R 3 whose spherical equation 
is p = cos 3<p. 

(d) Graph the surface in R 3 whose spherical equation 
is p = cos 38. 

41. (a) Graph the curve in R 2 whose polar equation is 

r = sin 38. 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = sin 38. 

(c) Graph the surface in R 3 whose spherical equation 
is p = sin 3(p. 



(d) Graph the surface in R 3 whose spherical equation 
is p = sin 38. Compare the results of this exercise 
with those of Exercise 40. 

42. (a) Graph the curve in R 2 whose polar equation is 

r — 1 + sin |. (This curve is known as a nephroid, 
meaning "kidney shaped.") 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = 1 + sin ~ . 

(c) Graph the surface in R 3 whose spherical equation 
is p = 1 + sin | . 

(d) Graph the surface in R 3 whose spherical equation 
is p = 1 + sin | . 

43. (a) Graph the curve in R 2 whose polar equation is 

r = 8. 

(b) Graph the surface in R 3 whose cylindrical equation 
is r = 8. 

(c) Graph the surface in R 3 whose spherical equation 
is p = <p, 

(d) Graph the surface in R 3 whose spherical equation 
is p = 8, where jt/2 < cp < n and 0 < 8 < 4n. 

44. Consider the solid hemisphere of radius 5 pictured in 
Figure 1.125. 

(a) Describe this solid, using spherical coordinates. 

(b) Describe this solid, using cylindrical coordinates. 



z 




y 



Figure 1.125 The solid hemisphere of 
Exercise 44. 



45. Consider the solid cylinder pictured in Figure 1.126. 

(a) Describe this solid, using cylindrical coordinates 
(position the cylinder conveniently). 

(b) Describe this solid, using spherical coordinates. 

h 6 - 




Figure 1.126 The solid 
cylinder of Exercise 45. 




2.1 Functions of Several 
Variables; Graphing 
Surfaces 

2.2 Limits 

2.3 The Derivative 

2.4 Properties; Higher-order 
Partial Derivatives 

2.5 The Chain Rule 

2.6 Directional Derivatives and 
the Gradient 

2.7 Newton's Method (optional) 

True/False Exercises for 
Chapter 2 

Miscellaneous Exercises for 
Chapter 2 



/ 




X Y 

Figure 2.1 The mapping 
nature of a function. 



Differentiation in 
Several Variables 



2.1 Functions of Several Variables; 
Graphing Surfaces 

The volume and surface area of a sphere depend on its radius, the formulas 
describing their relationships being V = |:rr 3 and S = 4nr 2 . (Here V and S 
are, respectively, the volume and surface area of the sphere and r its radius.) 
These equations define the volume and surface area as functions of the radius. 
The essential characteristic of a function is that the so-called independent variable 
(in this case the radius) determines a unique value of the dependent variable ( V 
or S). No doubt you can think of many quantities that are determined uniquely 
not by one variable (as the volume of a sphere is determined by its radius) but 
by several: the area of a rectangle, the volume of a cylinder or cone, the average 
annual rainfall in Cleveland or the national debt. Realistic modeling of the world 
requires that we understand the concept of a function of more than one variable 
and how to find meaningful ways to visualize such functions. 

Definitions, Notation, and Examples 

A function, any function, has three features: (1) a domain set X, (2) a codomain 
set Y, and (3) a rule of assignment that associates to each element x in the 
domain X a unique element, usually denoted f(x), in the codomain Y. We will 
frequently use the notation f:X—> Y for a function. Such notation indicates all 
the ingredients of a particular function, although it does not make the nature of 
the rule of assignment explicit. This notation also suggests the "mapping" nature 
of a function, indicated by Figure 2.1. 

EXAMPLE 1 Abstract definitions are necessary, but it is just as important that 
you understand functions as they actually occur. Consider the act of assigning to 
each U.S. citizen his or her social security number. This pairing defines a function: 
Each citizen is assigned one social security number. The domain is the set of U.S. 
citizens and the codomain is the set of all nine-digit strings of numbers. 

On the other hand when a university assigns students to dormitory rooms, it 
is unlikely that it is creating a function from the set of available rooms to the set of 
students. This is because some rooms may have more than one student assigned 
to them, so that a particular room does not necessarily determine a unique student 
occupant. ♦ 



2.1 | Functions of Several Variables; Graphing Surfaces 



DEFINITION 1 .1 The range of a function f:X ^ Y is the set of those 
elements of Y that are actual values of /. That is, the range of / consists of 
those y in Y such that y = f(x) for some x in X. 
Using set notation, we find that 

Range / = {y e Y \ y = f(x) for some x e X] . 



In the social security function of Example 1, the range consists of those nine- 
digit numbers actually used as social security numbers. For example, the number 
000-00-0000 is not in the range, since no one is actually assigned this number. 




Figure 2.2 Every y e Y is "hit" 
by at least one x e X. 




Figure 2.3 The element b e Y 
is not the image of any x £ X. 



DEFINITION 1 .2 A function /: X -> Y is said to be onto (or surjective) 

if every element of Y is the image of some element of X, that is, if range 
f=Y. 



The social security function is not onto, since 000-00-0000 is in the codomain 
but not in the range. Pictorially, an onto function is suggested by Figure 2.2. A 
function that is not onto looks instead like Figure 2.3. You may find it helpful to 
think of the codomain of a function / as the set of possible (or allowable) values 
of /, and the range of / as the set of actual values attained. Then an onto function 
is one whose possible and actual values are the same. 



DEFINITION 1 .3 A function /: X Y is called one-one (or injective) if 

no two distinct elements of the domain have the same image under /. That 
is, / is one-one if whenever x\, * 2 £ X and x\ ^ x 2 , then f{x\) ^ f(xz). 
(See Figure 2.4.) 




not one-one 



Figure 2.4 The figure on the left depicts a one-one mapping; the one 
on the right shows a function that is not one-one. 



One would expect the social security function to be one-one, but we have heard 
of cases of two people being assigned the same number so that, alas, apparently 
it is not. 

When you studied single-variable calculus, the functions of interest were 
those whose domains and codomains were subsets of R (the real numbers). It 
was probably the case that only the rule of assignment was made explicit; it is 
generally assumed that the domain is the largest possible subset of R for which 
the function makes sense. The codomain is generally taken to be all of R. 



84 Chapter 2 I Differentiation in Several Variables 



EXAMPLE 2 Suppose /: R —> R is given by /(jc) = x 2 . Then the domain and 
codomain are, explicitly, all of R, but the range of / is the interval [0, oo). Thus 
/ is not onto, since the codomain is strictly larger than the range. Note that / is 
not one-one, since f(2) = /(— 2) = 4, but 2 —2. ♦ 



EXAMPLE 3 Suppose g is a function such that g(x) = >Jx — 1. Then if we 
take the codomain to be all of R, the domain cannot be any larger than [1, oo). 
If the domain included any values less than one, the radicand would be negative 
and, hence, g would not be real-valued. ♦ 

Now we're ready to think about functions of more than one real variable. In 
the most general terms, these are the functions whose domains are subsets X of 
R" and whose codomains are subsets of R"', for some positive integers n and m. 
(For simplicity of notation, we'll take the codomains to be all of R'", except when 
specified otherwise.) That is, such a function is a mapping f:XQ R" R™ that 
associates to a vector (or point) x in X a unique vector (point) f(x) in R" ! . 



EXAMPLE 4 Let T: R 3 -* R be defined by T(x, y, z) = xy + xz + yz. We 
can think of T as a sort of "temperature function." Given a point x = (x, y, z) in 
R 3 , T(x) calculates the temperature at that point. ♦ 



EXAMPLE 5 Let L: R" -> R be given by L(x) = ||x|| . This is a "length func- 
tion" in that it computes the length of any vector x in R". Note that L is not 
one-one, since L(e,-) = L(e y ) = 1, where e, and e ; - are any two of the standard 
basis vectors for R" . L also fails to be onto, since the length of a vector is always 
nonnegative. ♦ 



EXAMPLE 6 Consider the function given by N(x) = x/||x|| where x is a vector 
in R 3 . Note that N is not defined if x = 0, so the largest possible domain for N is 
R 3 — {0}. The range of N consists of all unit vectors in R 3 . The function N is the 
"normalization function," that is, the function that takes a nonzero vector in R 3 
and returns the unit vector that points in the same direction. ♦ 



EXAMPLE 7 Sometimes a function may be given numerically by a table. 
One such example is the notion of windchill — the apparent temperature one 
feels when taking into account both the actual air temperature and the speed of 
the wind. A standard table of windchill values is shown in Figure 2.5. 1 From it 
we see that if the air temperature is 20 °F and the windspeed is 25 mph, the wind- 
chill temperature ("how cold it feels") is 3 °F. Similarly, if the air temperature is 
35°F and the windspeed is 10 mph, then the windchill is 27°F. In other words, 
if s denotes windspeed and t air temperature, then the windchill is a function 
W(s, t). ♦ 

The functions described in Examples 4, 5, and 7 are scalar-valued functions, 
that is, functions whose codomains are R or subsets of R. Scalar- valued functions 
are our main concern for this chapter. Nonetheless, let's look at a few examples 
of functions whose codomains are R" ! where m > 1 . 



NOAA, National Weather Service, Office of Climate, Water, and Weather Services, "NWS Wind Chill 
Temperature Index." February 26, 2004. <http://www.nws.noaa.gov/om/windchill> (July 31, 2010). 



2.1 | Functions of Several Variables; Graphing Surfaces 



Air Temp 
(deg F) 


Windspeed (mph) 


5 


10 


15 


20 


25 


30 


35 40 


45 


50 


55 


60 


40 


36 


34 


32 


30 


29 


28 


28 27 


26 


26 


25 


25 


35 


31 


27 


25 


24 


23 


22 


21 20 


19 


19 


18 


17 


30 


25 


21 


19 


17 


16 


15 


14 13 


12 


12 


11 


10 


25 


19 


15 


13 


11 


9 


8 


7 6 


5 


4 


4 


3 


20 


13 


9 


6 


4 


3 


1 


0 -1 


-2 


-3 


-3 


-4 


15 


7 


3 


0 


-2 


-4 


-5 


-7 -8 


-9 


-10 


-11 


-11 


10 


1 


-4 


-7 


-9 


-11 


-12 


-14 -15 


-16 


-17 


-18 


-19 


5 


-5 


-10 


-13 


-15 


-17 


-19 


-21 -22 


-23 


-24 


-25 


-26 


0 


-11 


-16 


-19 


-22 


-24 


-26 


-27 -29 


-30 


-31 


-32 


-33 


-5 


-16 


-22 


-26 


-29 


-31 


-33 


-34 -36 


-37 


-38 


-39 


-40 


-10 


-22 


-28 


-32 


-35 


-37 


-39 


-41 -43 


-44 


-45 


-46 


-48 


-15 


-28 


-35 


-39 


-42 


-44 


-46 


-48 -50 


-51 


-52 


-54 


-55 


-20 


-34 


-41 


-45 


-48 


-51 


-53 


-55 -57 


-58 


-60 


-61 


-62 


-25 


-40 


-47 


-51 


-55 


-58 


-60 


-62 -64 


-65 


-67 


-68 


-69 


-30 


-46 


-53 


-58 


-61 


-64 


-67 


-69 -71 


-72 


-74 


-75 


-76 


-35 


-52 


-59 


-64 


-68 


-71 


-73 


-76 -78 


-79 


-81 


-82 


-84 


-40 


-57 


-66 


-71 


-74 


-78 


-80 


-82 -84 


-86 


-88 


-89 


-91 


-45 


-63 


-72 


-77 


-81 


-84 


-87 


-89 -91 


-93 


-95 


-97 


-98 




Figure 2.6 The helix of 
Example 8. The arrow shows the 
direction of increasing t . 



Figure 2.5 Table of windchill values in English units. 

EXAMPLE 8 Define f: R -> R 3 by f(f) = (cos t, sin?, t). The range off is the 
curve in R 3 with parametric equations x = cos t , y = sin t , z = t . If we think of 
t as a time parameter, then this function traces out the corkscrew curve (called a 
helix) shown in Figure 2.6. ♦ 

EXAMPLE 9 We can think of the velocity of a fluid as a vector in R 3 . This 
vector depends on (at least) the point at which one measures the velocity and also 
the time at which one makes the measurement. In other words, velocity may be 
considered to be a function v: X c R 4 — >• R 3 . The domain X is a subset of R 4 
because three variables x, y, z are required to describe a point in the fluid and a 
fourth variable t is needed to keep track of time. (See Figure 2.7.) For instance, 
such a function v might be given by the expression 



v(x, y, z, t) = xyzt\ + (x 1 



y 2 )l 



(3z + /)k. 




Figure 2.7 A water 
pitcher. The velocity v of 
the water is a function 
from a subset of R 4 to R 3 . 



You may have noted that the expression for v in Example 9 is considerably 
more complicated than those for the functions given in Examples 4-8. This is 
because all the variables and vector components have been written out explicitly. 
In general, if we have a function f: X c R" — > R m , then xeX can be written as 
x = (xi , X2, ■ ■ ■ , x„) and f can be written in terms of its component functions 
/i. fz< ■ ■ ■ . fm- The component functions are scalar- valued functions of x £ X 
that define the components of the vector f(x) £ R" ! . What results is a morass of 
symbols: 

f(x) = f(jti, X2, ■ . ■ , x n ) (emphasizing the variables) 

— (/i( x )> /2( x )> ■ • • > /m( x )) (emphasizing the component functions) 
= {fi{x\,x 2 , . . .,x n ), f 2 (xi,x 2 , . . . ,X n ), f m (Xl,X 2 , ■ ■ .,x„)) 

(writing out all components). 



86 Chapter 2 I Differentiation in Several Variables 



For example, the function L of Example 5, when expanded, becomes 



L(x) = L(x\,x 2 , . . . , x„) = ^Jxf + x\ 
The function N of Example 6 becomes 

X (X],X2,X-i) 

N(x) = — 
^ llxll 




x 2 x 3 



and, hence, the three component functions of N are 

X\ X2 

Ni(xi,x 2 ,X3) = N 2 (xi, x 2 , xi) = 



X^ ~\~ X2 ~\~ X^ J X\ -\- x 2 ~\~ x^ 

x 3 

Ni(xi,X2, x 3 ) = 



X\ -\- Xj ~"T~ X-i 



Although writing a function in terms of all its variables and components has 
the advantage of being explicit, quite a lot of paper and ink are used in the process. 
The use of vector notation not only saves space and trees but also helps to make 
the meaning of a function clear by emphasizing that a function maps points in 
R" to points in R'" . Vector notation makes a function of 300 variables look "just 
like" a function of one variable. Try to avoid writing out components as much as 
you can (except when you want to impress your friends). 



Visualizing Functions 

No doubt you have been graphing scalar-valued functions of one variable for so 
long that you give the matter little thought. Let's scrutinize what you've been do- 
ing, however. A function /:XcR^ R takes a real number and returns another 
real number as suggested by Figure 2.8. The graph of / is something that "lives" 
in R 2 . (See Figure 2.9.) It consists of points (x, y) such that y = f(x). That is, 

Graph / = {{x, f(x)) \ x e X} = {(x, y)\xeX,y = f(x)} . 

The important fact is that, in general, the graph of a scalar-valued function 
of a single variable is a curve — a one-dimensional object — sitting inside two- 
dimensional space. 



y 



f 



SKA 




(x,f(x)) / 




X 



Figure 2.8 A function /:ICR->R, Figure 2.9 The graph off. 



2.1 | Functions of Several Variables; Graphing Surfaces 



Now suppose we have a function /:XcR 2 ->R, that is, a function of two 
variables. We make essentially the same definition for the graph: 



Graph / = {(x, /(x)) xel). (1) 



Of course, x = (x, y) is a point of R 2 . Thus, {(x, fix))} may also be written as 

l(x, y, fix, y))} , or as {(x, y, z) | (x, y) e X, z = f(x, y)} . 

Hence, the graph of a scalar-valued function of two variables is something that 
sits in R 3 . Generally speaking, the graph will be a surface. 

EXAMPLE 1 0 The graph of the function 

9 1,1,7 
/:R 2 ^R /(x,y)=-y 3 -y--x 2 + - 

is shown in Figure 2.10. For each point x = (x, y) in R 2 , the point in R 3 with 
coordinates (x, y, ^y 3 — y — \x 2 + |) is graphed. ♦ 




Figure 2.10 The graph of fix, y) = -j^y 3 — y — \x 2 + |. 



Graphing functions of two variables is a much more difficult task than graph- 
ing functions of one variable. Of course, one method is to let a computer do 
the work. Nonetheless, if you want to get a feeling for functions of more than 
one variable, being able to sketch a rough graph by hand is still a valuable skill. 
The trick to putting together a reasonable graph is to find a way to cut down on 
the dimensions involved. One way this can be achieved is by drawing certain 
special curves that lie on the surface z = fix, y). These special curves, called 
contour curves, are the ones obtained by intersecting the surface with horizontal 
planes z = c for various values of the constant c. Some contour curves drawn 
on the surface of Example 10 are shown in Figure 2.1 1. If we compress all the 
contour curves onto the xy-plane (in essence, if we look down along the posi- 
tive z-axis), then we create a "topographic map" of the surface that is shown in 
Figure 2.12. These curves in the xy-plane are called the level curves of the original 
function /. 

The point of the preceding discussion is that we can reverse the process in 
order to sketch systematically the graph of a function / of two variables: We 



88 Chapter 2 I Differentiation in Several Variables 




Figure 2.1 1 Some contour curves of the Figure 2.1 2 Some level curves of 

function in Example 10. the function in Example 10. 



first construct a topographic map in R 2 by finding the level curves of /, then 
situate these curves in R 3 as contour curves at the appropriate heights, and finally 
complete the graph of the function. Before we give an example, let's restate our 
terminology with greater precision. 



DEFINITION 1 .4 Let /:XcR 2 ^Rbea scalar-valued function of two 
variables. The level curve at height c of / is the curve in R 2 defined by the 
equation f(x, y) = c, where c is a constant. In mathematical notation, 

Level curve at height c = {(x, y) £ R 2 | fix, y) = c] . 

The contour curve at height c of / is the curve in R 3 defined by the two 
equations z = fix, y) and z = c. Symbolized, 

Contour curve at height c = {(x, y, z) £ R 3 | z = fix, y) = c} . 

In addition to level and contour curves, consideration of the sections of a 
surface by the planes where x or y is held constant is also helpful. A section of a 
surface by a plane is just the intersection of the surface with that plane. Formally, 
we have the following definition: 



DEFINITION 1 .5 Let /: X C R 2 -> R be a scalar-valued function of two 
variables. The section of the graph of / by the plane x = c (where c 
is a constant) is the set of points (x, y, z), where z = f(x, y) and x = c. 
Symbolized, 

Section by x = c is {(x, y, z) e R 3 I z = f(x, y), x = c}. 

Similarly, the section of the graph of / by the plane y = c is the set of 

points described as follows: 

Section by y = c is {(x, y, z) e R 3 | z = fix, y), y = c]. 



2.1 | Functions of Several Variables; Graphing Surfaces 



EXAMPLE 1 1 We'll use level and contour curves to construct the graph of the 
function 

/:R 2 ^R, f(x,y) = 4-x 2 -y 2 . 

By Definition 1 .4, the level curve at height c is 

{(x, y) 6 R 2 | 4 - x 2 - y 2 = c} = {(x, y) | x 2 + y 2 = 4 - c} . 

Thus, we see that the level curves for c < 4 are circles centered at the origin of 
radius -J4 — c. The level "curve" at height c = 4 is not a curve at all but just a 
single point (the origin). Finally, there are no level curves at heights larger than 4 
since the equation x 2 + y 2 = 4 — c has no real solutions in x and y. (Why not?) 
These remarks are summarized in the following table: 



c 


Level curve x 2 + y 2 = 4 — c 


-5 


x 2 + y 2 = 9 


-1 


x 2 + y 2 = 5 


0 


x 2 + y 2 = 4 


1 


x 2 + y 2 = 3 


3 


x 2 + y 2 = l 


4 


x 2 + y 2 = 0 x = y = 0 


c, where c > 4 


empty 



Thus, the family of level curves, the "topographic map" of the surface z = 
4 — x 2 — y 2 , is shown in Figure 2.13. Some contour curves, which sit in R 3 , 
are shown in Figure 2.14, where we can get a feeling for the complete graph of 
z = 4 — x 2 — y 2 . It is a surface that looks like an inverted dish and is called a 
paraboloid. (See Figure 2. 15.) To make the picture clearer, we have also sketched 
in the sections of the surface by the planes x = 0 and y = 0. The section by x = 0 
is given analytically by the set 

{(x, y, z) e R 3 | z = 4 - x 2 - y 2 , x = 0} = {(0, y, z )\z = 4- y 2 } . 

Similarly, the section by y = 0 is 

{(x, y, z) e R 3 | z = 4 - x 2 - y 2 , y = 0} = {(x, 0, z) \ z = 4 - x 2 } . 




Figure 2.13 The topographic Figure 2.14 Some contour Figure 2.15 The graph of 

map of z = 4 — x 2 — y 2 (i.e., curves of z = 4 — x 2 — y 2 . f(x, y) = 4 — x 2 — y 2 . 

several of its level curves). 



Chapter 2 | Differentiation in Several Variables 



Since these sections are parabolas, it is easy to see how this surface obtained its 
name. ♦ 

EXAMPLE 12 We'll graph the function g: R 2 -> R, g(x, y) = y 2 - x 2 . The 
level curves are all hyperbolas, with the exception of the level curve at height 0, 
which is a pair of intersecting lines. 



c = -4 






c 


Level curve y 2 — x 2 = c 




-4 




x 2 - y 2 = 4 




4 


-1 




x 2 - y 2 = 1 






0 


y 2 -X 2 =0 


<S=> (y-x)(y+x) = 0 J= 


=>■ y = ±x 


X 


1 




y 2 -x 2 = \ 






4 




y 2 -X 2 =4 





Figure 2.1 6 Some level curves 
of g(x,y) = y 2 -x 2 . 




Figure 2.1 7 The contour curves 
and graph of g(x, y) = y 2 — x 2 . 



The collection of level curves is graphed in Figure 2.16. The sections by x = c 
are 

l(x, y, z) | z = y 2 - x 2 ,x = c} = {(c, y, z) \ z = y 2 - c 2 }. 

These are clearly parabolas in the planes x = c. The sections by y = c are 

{(x, y, z) | z = y 2 - x 2 , y = c) = {(c, y, z) | z = c 2 - x 2 }, 

which are again parabolas. The level curves and sections generate the contour 
curves and surface depicted in Figure 2.17. Perhaps understandably, this surface 
is called a hyperbolic paraboloid. ♦ 

EXAMPLE 13 We compare the graphs of the function f(x, y) = 4 — x 2 — y 2 
of Example 1 1 with that of 

h: R 2 - {(0, 0)} -> R, h(x, y) = ln(x 2 + y 2 ). 

The level curve of h at height c is 

{(jc, y) 6 R 2 | ln(x 2 + y 2 ) = c] = {(x, y) | x 2 + y 2 = e c \ . 

Since e c > 0 for all c 6 R, we see that the level curve exists for any c and is a 
circle of radius J~e^ = e c/2 . 



c 


Level curve x 2 + y 2 = e c 


-5 


x 2 + y 2 = e~ 5 


-1 


x 2 + y 2 = e~ l 


0 


x 2 + y 2 = 1 


1 


x 2 + y 2 = e 


3 


x 2 + y 2 = e 3 


4 


x 2 +y 2 = e 4 



The collection of level curves is shown in Figure 2.18 and the graph in Figure 2.19. 
Note that the section of the graph by x = 0 is 

{(x,y,z) eR 3 | Z = \n(x 2 + y 2 ),x = 0} = {(0, y, z) \ z = ln(y 2 ) = 2 In | y | } . 



2.1 | Functions of Several Variables; Graphing Surfaces 



y 




X 



Figure 2.1 8 The collection of level curves ofz = ln(x 2 + v 2 ). 



The section by _y = 0 is entirely similar: 

{(x,y,z)eR 3 | z = ln(x 2 + y 2 ), y = 0} = {(*,(>, z)| z = ln(x 2 ) = 2 In \x\ } . 

♦ 

In fact, if we switch from Cartesian to cylindrical coordinates, it is quite 
easy to understand the surfaces in both Examples 11 and 13. In view of the 
Cartesian/cylindrical relation x 2 + y 2 = r 2 , we see that for the function / of 
Example 11, 

z = 4 - x 2 - y 2 = 4 - (x 2 + y 2 ) = 4 - r 2 . 

For the function h of Example 13, we have 

z = ln(x 2 + y 2 ) = ln(r 2 ) = 21nr, 

where we assume the usual convention that the cylindrical coordinate r is non- 
negative. Thus both of the graphs in Figures 2.15 and 2.19 are of surfaces of 
revolution obtained by revolving different curves about the z-axis. As a result, 
the level curves are, in general, circular. 

The preceding discussion has been devoted entirely to graphing scalar- valued 
functions of just two variables. However, all the ideas can be extended to more 
variables and higher dimensions. If /: X c R" —> R is a (scalar-valued) function 
of n variables, then the graph of / is the subset of R" +1 given by 



Graph / = {(x, /(x)) xel} 






= {{x\, ■ ■ ■ , X„, X n+ \) 


(Xi, . . .,X n ) € X, 


(2) 




x n+\ = f( x l , ■ ■ ■ , X n )} . 





92 Chapter 2 | Differentiation in Several Variables 



z 




y 



Figure 2.20 The level sets of the 
function F(x, y,z) = x + y + z 
are planes in R 3 . 



y 




(1,0) 

i x 



Figure 2.21 The unit circle 
x 2 + y 2 = 1. 



Z 




Figure 2.22 The sphere of radius 
a, centered at (xq, yo, zo). 



(The compactness of vector notation makes the definition of the graph of a function 
of n variables exactly the same as in ( 1 ).) The level set at height c of such a function 
is defined by 

Level set at height c = {x e R" | f(x) = c} 

= {(*i, X 2 , . . . , X„) | f(Xl, X2, ...,X») = C}. 

While the graph of / is a subset of R" +1 , a level set of / is a subset of R". This 
makes it possible to get some geometric insight into graphs of functions of three 
variables, even though we cannot actually visualize them. 

EXAMPLE 14 Let F: R 3 -> R be given by F(x, y, z) = x + y + z. Then the 
graph of F is the set {(*, y, z, if) | w = x + y + z} and is a subset (called a 
hypersurface) of R 4 , which we cannot depict adequately. Nonetheless, we can 
look at the level sets of F, which are surfaces in R 3 . (See Figure 2.20.) We have 

Level set at height c = {(x, y, z) \ x + y + z = c}. 

Thus, the level sets form a family of parallel planes with normal vector i + j + k. 

♦ 

Surfaces in General 

Not all curves in R 2 can be described as the graph of a single function of one 
variable. Perhaps the most familiar example is the unit circle shown in Figure 2.2 1 . 
Its graph cannot be determined by a single equation of the form y = f(x) (or, for 
that matter, by one of the form x = g(y)). As we know, the graph of the circle may 
be described analytically by the equation x 2 + y 2 = 1. In general, a curve in R 2 
is determined by an arbitrary equation in x and y, not necessarily one that isolates 
y alone on one side. In other words, this means that a general curve is given by an 
equation of the form F(x, y) — c (i.e., a level set of a function of two variables). 

The analogous situation occurs with surfaces in R 3 . Frequently a surface is 
determined by an equation of the form F(x, y, z) — c (i.e., as a level set of a 
function of three variables), not necessarily one of the form z = f(x, y). 

EXAMPLE 15 A sphere is a surface in R 3 whose points are all equidistant from 
a fixed point. If this fixed point is the origin, then the equation for the sphere is 

llx-OH = |WI =o, (3) 

where a is a positive constant and x = (x, y, z) is a point on the sphere. If we 
square both sides of equation (3) and expand the (implicit) dot product, then we 
obtain perhaps the familiar equation of a sphere of radius a centered at the origin: 

x 2 + y 2 + Z 2 = a 2 . (4) 

If the center of the sphere is at the point xo = (jto, yo, zq), rather than the origin, 
then equation (3) should be modified to 

l|x-x 0 ||=a. (5) 

(See Figure 2.22.) 

When equation (5) is expanded, the following general equation for a sphere 
is obtained: 



(x - x 0 ) 2 + (y- y 0 ) 2 + (z - zo) 2 = a 2 (6) 



2.1 | Functions of Several Variables; Graphing Surfaces 



In the equation for a sphere, there is no way to solve for z uniquely in terms 
of x and y. Indeed, if we try to isolate z in equation (4), then 

2 2 2 2 

z = a — x — y , 

so we are forced to make a choice of positive or negative square roots in order to 
solve for z: 

z = \J a 1 — x 2 — y 2 or z. = —\J a 2 — x 2 — y 2 . 

The positive square root corresponds to the upper hemisphere and the negative 
square root to the lower one. In any case, the entire sphere cannot be the graph 
of a single function of two variables. ♦ 

Of course, the graph of a function of two variables does describe a surface in 
the "level set" sense. If a surface happens to be given by an equation of the form 

z = f(x, y) 

for some appropriate function /:XcR 2 ^R, then we can move z to the oppo- 
site side, obtaining 

f{x, y)-z = 0. 
If we define a new function F of three variables by 

F{x,y,z) = f(x,y)-z, 

then the graph of / is precisely the level set at height 0 of F. We reiterate this 
point since it is all too often forgotten: The graph of a function of two variables is 
a surface in R 3 and is a level set of a function of three variables. However, not all 
level sets of functions of three variables are graphs of functions of two variables. 
We urge you to understand this distinction. 

Quadric Surfaces 

Conic sections, those curves obtained from the intersection of a cone with various 
planes, are among the simplest, yet also the most interesting, of plane curves: 
They are the circle, the ellipse, the parabola, and the hyperbola. Besides being 
produced in a similar geometric manner, conic sections have an elegant algebraic 
connection: Every conic section is described analytically by a polynomial equation 
of degree two in two variables. That is, every conic can be described by an equation 
that looks like 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 

for suitable constants A, . . . , F. 

In R 3 , the analytic analogue of the conic section is called a quadric surface. 
Quadric surfaces are those defined by equations that are polynomials of degree 
two in three variables: 

Ax 2 + Bxy + Cxz + Dy 2 + Eyz + Fz 2 + Gx + Hy + lz + J = 0. 

To pass from this equation to the appropriate graph is, in general, a cumbersome 
process without the aid of either a computer or more linear algebra than we 
currently have at our disposal. So, instead we offer examples of those quadric 
surfaces whose axes of symmetry lie along the coordinate axes in R 3 and whose 
corresponding analytic equations are relatively simple. In the discussion that 
follows, a, b, and c are constants, which, for convenience, we take to be positive. 



94 Chapter 2 | Differentiation in Several Variables 



Z 





Figure 2.26 The elliptic 



cone 




Figure 2.27 The graph of the 

2 2 2 

x y z 
equation — r + -pr T = 1 is a 

a 2 - b L c L 

hyperboloid of one sheet. 



Ellipsoid (Figure 2.23.) x 2 /a 2 + y 2 /b 2 + z 2 /c 2 = 1. 
This is the three-dimensional analogue of an ellipse in the plane. The sections 
of the ellipsoid by planes perpendicular to the coordinate axes are all ellipses. 
For example, if the ellipsoid is intersected with the plane z = 0, one obtains the 
standard ellipse x 2 /a 2 + y 2 /b 2 = 1, z = 0. If a = b = c, then the ellipsoid is a 
sphere of radius a. 

Elliptic paraboloid (Figure 2.24.) z/c = x 2 /a 2 + y 2 /b 2 . 

(The roles of x, y, and z may be interchanged.) This surface is the graph of 
a function of x and y. The paraboloid has elliptical (or single-point or empty) 
sections by the planes "z = constant" and parabolic sections by "x = constant" or 
" v = constant" planes. The constants a and b affect the aspect ratio of the elliptical 
cross sections, and the constant c affects the steepness of the dish. (Larger values 
of c produce steeper paraboloids.) 

Hyperbolic paraboloid (Figure 2.25.) z/c = y 2 /b 2 — x 2 /a 2 . 

(Again the roles of x, y, and z may be interchanged.) We saw the graph of this 
surface earlier in Example 12 of this section. It is shaped like a saddle whose "x = 
constant" or "y = constant" sections are parabolas and "z = constant" sections 
are hyperbolas. 

Elliptic cone (Figure 2.26.) z 2 /c 2 = x 2 /a 2 + y 2 /b 2 . 

The sections by "z = constant" planes are ellipses. The sections by x = 0 or 

y = 0 are each a pair of intersecting lines. 

Hyperboloid of one sheet (Figure 2.27.) x 2 /a 2 + y 2 /b 2 - z 2 /c 2 = 1. 

The term "one sheet" signifies that the surface is connected (i.e., that you can 
travel between any two points on the surface without having to leave the surface). 
The sections by "z = constant" planes are ellipses and those by "x = constant" 
or "y = constant" planes are hyperbolas, hence, this surface's name. 

Hyperboloid of two sheets (Figure 2.28.) z 2 /c 2 - x 2 /a 2 - y 2 /b 2 = 1. 

The fact that the left-hand side of the defining equation is the opposite of the left 
side of the equation for the previous hyperboloid is what causes this surface to 
consist of two pieces instead of one. More precisely, consider the sections of the 



2.1 | Exercises 95 



(0, 0, c) 




Figure 2.28 The graph of 
the equation 

2 2 2 

z x y 

c L or b L 
hyperboloid of two sheets. 




Figure 2.29 The 

hyperboloids 

asymptotic to the cone 
c 2 ~ a 2 + b 2 ' 



surface by planes of the form z = k for different constants k. These sections are 
thus given by 



or, equivalently, by 



7i 



2 n 
X V 

I- — 

a 2 b 2 



r 
b 2 

e 



= 1, 



z = k 



1, z = k. 



If —c < k < c, then 0 < k 2 /c 2 < 1. Thus, /: 2 /c 2 — 1 < 0, and so the preceding 
equation has no solution in x and y. Hence, the section by z = k, where \k\ < c, 
is empty. If \k\ > c, then the section is an ellipse. The sections by "x = constant" 
or "y = constant" planes are hyperbolas. 



In the same way that the hyperbolas 

2 n 

x y 
a 2 b 2 

are asymptotic to the lines y = ±{b/a)x, the hyperboloids 



= ±1 



2 2 1 

Z _ x y L 
c 2 a 2 b 2 



± 1 



are asymptotic to the cone 



= X -, + y 



b 2 ' 



This is perhaps intuitively clear from Figure 2.29, but let's see how to prove it 
rigorously. In our present context, to say that the hyperboloids are asymptotic 
to the cones means that they look more and more like the cones as |z| becomes 
(arbitrarily) large. Analytically, this should mean that the equations for the hy- 
perboloids should approximate the equation for the cone for sufficiently large \z\. 
The equations of the hyperboloids can be written as follows: 

r 2 v 2 7 2 7 2 / r 1 

x . y * ±1 * (i±* 



a 2 b 2 c A c- \ z A 

As \z\ -> oo, c 2 /z 2 -> 0, so the right side of the equation for the hyperboloids 
approaches z 2 /c 2 . Hence, the equations for the hyperboloids approximate that of 
the cone, as desired. 



2.1 Exercises 



1. Let/:R^ R be given by f(x) = 2x 2 + 1. 

(a) Find the domain and range of /. 

(b) Is / one-one? 

(c) Is / onto? 

2. Let g: R 2 -> Rbe given by g(x, y) = 2x 2 + 3y 2 — 7. 

(a) Find the domain and range of g. 

(b) Find a way to restrict the domain to make a new 
function with the same rule of assignment as g that 
is one-one. 



(c) Find a way to restrict the codomain to make a new 
function with the same rule of assignment as g that 
is onto. 

Find the domain and range of each of the functions given in 
Exercises 3-7. 

3. fix, y)=- 

y 

4. f(x, y) = ln(x + y) 

5. g(x, y, z) = Jx 2 + (y- 2) 2 + (z + I) 2 



Chapter 2 | Differentiation in Several Variables 



6. g(x, y, z) = 



7. f(x,y)= [x + y 



y-1 



x 2 +y 2 



8. Let f:R 2 -> R 3 be defined by f(x, y) = (x + y, 
ye x , x 2 y + 7). Determine the component functions 
off. 

9. Determine the component functions of the function v 
in Example 9. 

1 0. Let f : R 3 -> R 3 be defined by f(x) = x + 3 j . Write out 
the component functions of f in terms of the compo- 
nents of the vector x. 

1 1 . Consider the mapping that assigns to a nonzero vector 
x in R 3 the vector of length 2 that points in the direction 
opposite to x. 

(a) Give an analytic (symbolic) description of this 
mapping. 

(b) If x = (x, y, z), determine the component func- 
tions of this mapping. 

1 2. Consider the function f: R 2 —> R 3 given by f(x) = Ax, 
2 -1 " 

where A = 5 0 
-6 3 



and the vector x in R is 



written as the 2x1 column matrix x 



X] 
X2 



(a) Explicitly determine the component functions of 
f in terms of the components X\, X2 of the vector 
(i.e., column matrix) x. 

(b) Describe the range of f. 

1 3. Consider the function f: R 4 — > R 3 given by f(x) = Ax, 
~ 2 0 -1 1 
0 3 0 0 
2 0-11 



and the vector x in R 4 



is written as the 4x1 column matrix x 



x\ 

Xl 
-V3 
X4 



(a) Determine the component functions of f in terms 
of the components xi , xi, *3, X4 of the vector (i.e., 
column matrix) x. 

(b) Describe the range of f. 

In each of Exercises 14-23, (a) determine several level curves 
of the given function f (make sure to indicate the height c of 
each curve); (b) use the information obtained in part (a) to 
sketch the graph of f. 



14. f(x, y) = 3 

15. f(x,y) 

16. f(x,y) = x 2 + y 2 



x 2 + v 2 



17. 


f(x,y) = 




18. 


fix y) = 


Ax 2 


19. 


fix, y) = 


xy 


20. 


fix, y) = 


y 






X 


21. 


fix, y) = 


X 






y 


22. 


fix, y) = 


3 - 


23. 


fix, y) = 


\x\ 



In Exercises 24-27, use a computer to provide a portrait of 
the given function g(x, y). To do this, (a) use the computer to 
help you understand some of the level curves of the function, 
and (b) use the computer to graph (a portion of) the surface 
Z = g(x, y). In addition, mark on your surface some of the 
contour curves corresponding to the level curves you obtained 
in part (a). (See Figures 2.10 and 2.11.) 

^ 24. g(x, y) = ye x 

^ 25. g(x,y) = x 2 - xy 

^ 26. g(x, y) = ix 2 + 3y 2 y- x2 - vl 

^ sin(2 — x 2 — v 2 ) 

O 27. g{x, y) = \ . 

x £ + y L + 1 

28. The ideal gas law is the equation PV = kT, where P 
denotes the pressure of the gas, V the volume, T the 
temperature, and k is a positive constant. 

(a) Describe the temperature T of the gas as a function 
of volume and pressure. Sketch some level curves 
for this function. 

(b) Describe the volume V of the gas as a function 
of pressure and temperature. Sketch some level 
curves. 

29. (a) Graph the surfaces z = x 2 and z = y 2 ■ 

(b) Explain how one can understand the graph of the 
surfaces z = fix) and z = /(y) by considering 
the curve in the ifii-plane given by v = /(«). 

(c) Graph the surface in R 3 with equation y = x 2 . 

30. Use a computer to graph the family of level curves for 
the functions in Exercises 20 and 21 and compare your 
results with those obtained by hand sketching. How do 
you account for any differences? 

31 . Given a function f(x , y), can two different level curves 
of / intersect? Why or why not? 

In Exercises 32—36, describe the graph of g(x, y, z) by com- 
puting some level surfaces. (If you prefer, use a computer to 
assist you.) 

32. g(x, y,z) = x — 2y + 3z 

33. g(x, y,z) = x 2 + y 2 — z 



2.2 | Limits 



34. g(x, y, z ) = x 2 + y 2 + z 2 

35. g(x, y, z ) = x 2 + 9y 2 + 4z 2 

36. g(x, y, z) = xy - yz 

37. (a) Describe the graph of g(x, y, z) = x 2 + y 2 by 

computing some level surfaces. 

(b) Suppose g is a function such that the expres- 
sion for g(x, y, z) involves only x and y (i.e., 
g(x, y, z) = h(x, yj). What can you say about the 
level surfaces of g? 

(c) Suppose g is a function such that the expression 
for g(x, y, z) involves only x and z. What can you 
say about the level surfaces of g? 

(d) Suppose g is a function such that the expression 
for g(x, y, z) involves only x. What can you say 
about the level surfaces of g? 

38. This problem concerns the surface determined by the 
graph of the equation x 2 + xy — xz = 2. 

(a) Find a function F(x, y, z) of three variables so 
that this surface may be considered to be a level 
set of F. 

(b) Find a function f(x, y) of two variables so that 
this surface may be considered to be the graph of 
z = fix, y). 



42. x 



y 2 z 2 



39. Graph the ellipsoid 



x 2 y 2 2 



i. 



Is it possible to find a function f(x , y ) so that this ellip- 
soid may be considered to be the graph of z = f(x, y)? 
Explain. 

Sketch or describe the surfaces in R 3 determined by the equa- 
tions in Exercises 40—46. 



2 2 

43. x 2 + — - — = 0 
9 16 



40. z 



y 



41. z 2 



y 



x y i z" 

44. — + — = 1 

4 16 9 

x 2 y 2 - 

45. — + — = z 2 - 1 
25 16 

46. z = y 2 + 2 

We can look at examples of quadric surfaces with centers 
or vertices at points other than the origin by employing a 
change of coordinates of the form x = x — xq, y = y — y 0 , 
and z = Z — Zo- This coordinate change simply puts the point 
(xo, yo, Zo) of the xyz-coordinate system at the origin of the 
xyz-coordinate system by a translation of axes. Then, for ex- 
ample, the surface having equation 

(x - l) 2 (y + 2) 2 , 

can be identified by setting x = x — 1, y = y + 2, and z = 
Z — 5, so that we obtain 

-2 -2 

X V _ 2 

which is readily seen to be an ellipsoid centered at (1, —2, 5) 
of the xyz-coordinate system. By completing the square in x, 
y, or z as necessary, identify and sketch the quadric surfaces 
in Exercises 47—52. 

47. (* - l) 2 + (y + l) 2 = (z + 3) 2 

48. z = 4* 2 + (y + 2) 2 

49. Ax 2 + y 2 + z 2 + 8* = 0 

50. Ax 2 + y 2 - 4z 2 + 8x - 4y + 4 = 0 

51. * 2 + 2y 2 -6jt-z+10 = 0 

52. 9x 2 + 4y 2 - 36z 2 - 8y - 144z = 104 



2.2 Limits 

As you may recall, limit processes are central to the development of calculus. The 
mathematical and philosophical debate in the 1 8th and 1 9th centuries surrounding 
the meaning and soundness of techniques of taking limits was intense, questioning 
the very foundations of calculus. By the middle of the 19th century, the infamous 
"e — 8" definition of limits had been devised chiefly by Karl Weierstrass and 
Augustin Cauchy, much to the chagrin of many 20th (and 2 1 st) century students of 
calculus. In the ensuing discussion, we study both the intuitive and rigorous mean- 
ings of the limit of a function f : X C R" — > R m and how limits lead to the notion 
of a continuous function, our main object of study for the remainder of this text. 



98 Chapter 2 I Differentiation in Several Variables 



The Notion of a Limit 

For a scalar- valued function of a single variable, /:XcR->R, you have seen 
the statement 

lim f(x) = L 

and perhaps have an intuitive understanding of its meaning. In imprecise terms, 
the preceding equation (read "The limit of fix) as x approaches a is L.") means 
that you can make the numerical value of f(x) arbitrarily close to L by keeping 
x sufficiently close (but not equal) to a. This idea generalizes immediately to 
functions f:Xc R" — ► R m . In particular, by writing the equation 

lim f(x) = L, 

x^a 

where f:Xc R" R m , we mean that we can make the vector f(x) arbitrarily 
close to the limit vector L by keeping the vector x g X sufficiently close (but not 
equal) to a. 

The word "close" means that the distance (in the sense of § 1 .6) between f(x) 
and L is small. Thus, we offer a first definition of limit using the notation for 
distance. 



DEFINITION 2.1 (Intuitive definition of limit) The equation 

lim f(x) = L, 

x— »a 

where HcR"^ R m , means that we can make ||f(x) — L|| arbitrarily 
small (i.e., near zero) by keeping ||x — a|| sufficiently small (but nonzero). 



In the case of a scalar- valued function /:XcR" -> R, the vector length 
||f(x) — L|| can be replaced by the absolute value |/(x) — L\. Similarly, if / is a 
function of just one variable, then ||x — a|| can be replaced by \x — a\. 



EXAMPLE 1 Suppose that /: R -> R is given by 



1 



Figure 2.30 The graph of / of 
Example 1. 



/(*) = 



0 



ifx < 1 
if x > 1 



The graph of / is shown in Figure 2.30. What should lim v ^i f(x) be? The limit 
can't be 0, because no matter how near we make x to 1 (i.e., no matter how small 
we take \x — 1 1), the values of x can be both slightly larger and slightly smaller 
than 1 . The values of / corresponding to those values of x larger than 1 will 
be 2. Thus, for such values of x, we cannot make \f(x) — 0| arbitrarily small, 
since, for x > 1, \f(x) — 0| = |2 — 0| = 2. Similarly, the limit can't be 2, since 
no matter how small we take \x — l\,x can be slightly smaller than 1. For x < 1, 
f(x) = 0 and therefore, we cannot make \f(x) — 2| = |0 — 2| = 2 arbitrarily 
small. Indeed it should now be clear that the limit can't be L for any L e R. 
Hence, lim x ^i f(x) does not exist for this function. ♦ 



EXAMPLE 2 Let f: R 2 R 2 be defined by f(x) = 5x. (That is, f is five times 
the identity function.) Then it should be obvious intuitively that 



lim f(x) 

x^i+j 



lim 5x : 



5i + 5j. 



2.2 | Limits 



Indeed, if we write x = xi + yj, then 
||f(x)-(5i + 5j)|| = ||(5xi + 5yj)-(5i + 5j)|| 

= \\5 (x- l)i + 5(y- l)j || = V25(x- l) 2 + 25(j- l) 2 
= 5j(x - l) 2 + (y - 1)2. 

This last quantity can be made as small as we wish by keeping 

I|X " (i + j)|| = y/{x ~ l) 2 + (y - l) 2 

sufficiently small. ♦ 

EXAMPLE 3 Now suppose that g:R"^R" is defined by g(x) = 3x. We 
claim that, for any a e R", 

lim 3x = 3a. 

x^a 

In other words, we claim that ||3x — 3a || can be made as small as we like by 
keeping ||x — a|| sufficiently small. Note that 

||3x-3a|| = ||3(x-a)|| =3||x-a||. 

This means that if we wish to make ||3x — 3a|| no more than, say, 0.003, then 
we may do so by making sure that ||x — a|| is no more than 0.001. If, instead, 
we want ||3x — 3a|| to be no more than 0.0003, we can achieve this by keeping 
|| x — a|| no more than 0.0001. Indeed, if we want ||3x — 3a|| to be no more than 
any specified amount (no matter how small), then we can achieve this by making 
sure that ||x — a|| is no more than one-third of that amount. 

More generally, if h: R" —> R" is any constant k times the identity function 
(i.e., h(x) = kx) and a e R" is any vector, then 

lim h(x) = lim kx = ka. ♦ 

x^a x— >a 

The main difficulty with Definition 2.1 lies in the terms "arbitrarily small" 
and "sufficiently small." They are simply too vague. We can add some precision 
to our intuition as follows: Think of applying the function f: X c R" R" ! as 
performing some sort of scientific experiment. Letting the variable x take on 
a particular value in X amounts to making certain measurements of the input 
variables to the experiment, and the resulting value f(x) can be considered to be the 
outcome of the experiment. Experiments are designed to test theories, so suppose 
that this hypothetical experiment is designed to test the theory that as the input is 
closer and closer to a, then the outcome gets closer and closer to L. To verify this 
theory, you should establish some acceptable (absolute) experimental error for the 
outcome, say, 0.05. That is, you want ||f(x) — L|| < 0.05, if ||x — a|| is sufficiently 
small. Then just how small does ||x — a | need to be? Perhaps it turns out that you 
must have ||x — a|| < 0.02, and that if you do take ||x — a|| < 0.02, then indeed 
||f(x) — L|| < 0.05. Does this mean that your theory is correct? Not yet. Now, 
suppose that you decide to be more exacting and will only accept an experimental 
error of 0.005 instead of 0.05. In other words, you desire ||f(x) — L|| < 0.005. 
Perhaps you find that if you take ||x — a|| < 0.001, then this new goal can be 
achieved. Is your theory correct? Well, there's nothing sacred about the number 
0.005, so perhaps you should insist that ||f(x) — L|| < 0.001, or that ||f(x) — L|| < 
0.00001. The point is that if your theory really is correct, then no matter what 
(absolute) experimental error e you choose for your outcome, you should be 
able to find a "tolerance level" 8 for your input x so that if ||x — a|| < <5, then 



Chapter 2 | Differentiation in Several Variables 



|| f(x) — L || < e. It is this heuristic approach that motivates the technical definition 
of the limit. 



DEFINITION 2.2 (RIGOROUS DEFINITION OF limit) Let f : X C R" R m 

be a function. Then to say 

lim f(x) = L 

x— *a 

means that given any e > 0, you can find a S > 0 (which will, in general, 
depend on e) such that if x e X and 0 < ||x — a|| < <5, then ||f(x) — L|| < e. 



The condition 0 < ||x — a|| simply means that we care only about values 
f(x) when x is near a, but not equal to a. Definition 2.2 is not easy to use in 
practice (and we will not use it frequently). Moreover, it is of little value insofar 
as actually evaluating limits of functions is concerned. (The evaluation of the 
limit of a function of more than one variable is, in general, a difficult task.) 

EXAMPLE 4 So that you have some feeling for working with Definition 2.2, 
let's see rigorously that 

lim (3x - 5y + 2z) = 12 

tw)->- (1,-1,2) 

(as should be "obvious"). This means that given any number e > 0, we can find 
a corresponding S > 0 such that 

ifO < \\(x,y,z)-(l, -1,2)|| < 8, then \3x -5y + 2 z - 12| < e. 

(Note the uses of vector lengths and absolute values.) We'll present a formal 
proof in the next paragraph, but for now we'll do the necessary background 
calculations in order to provide such a proof. First, we need to rewrite the two 
inequalities in such a way as to make it more plausible that the e-inequality could 
arise algebraically from the S -inequality. From the definition of vector length, the 
(5-inequality becomes 

0 < y/(x- l) 2 + (y + l) 2 + (z-2) 2 < 8. 
If this is true, then we certainly have the three inequalities 

V(jc - l) 2 = \x - 1| <S, 

y ( V+ l)2 = |y + l| <S , 

y/(z - 2) 2 = \z-2\ < S. 

Now, rewrite the left side of the e-inequality and use the triangle inequality (2) 
of §1.6: 

\3x -5y + 2z- 12| = |3(* - 1) - 5(y + 1) + 2(z - 2)| 

< |3(x - 1)| + \5(y + 1)| + \2(z - 2)| 
= 3\x - 1| +5|y + 1| +2|z-2|. 

Thus, if 

0< \\(x,y,z)- (1,-1, 2)|| <S, 

then 

\x- 1| <8, \y+l\ <S, and \z-2\ < 8, 

so that 



\3x-5y + 2z- 12| < 3\x - 1| + 5\y + 1| + 2\z - 2\ 
< 38 + 58 + 28 = 108. 



2.2 | Limits 101 



If we think of S as a positive quantity that we can make as small as desired, then 
105 can also be made small. In fact, it is 10<5 that plays the role of e. 

Now for a formal, "textbook" proof: Given any e > 0, choose 8 > 0 so that 
S < e/10. Then, if 

0< \\(x,y,z)- (1,-1, 2)|| <S, 

it follows that 

|x-l|<<5, |y+l|<<5, and \z-2\<8, 

so that 

\3x-5y + 2z- 12| < 3|x - 1| + 5\y + 1|+ 2|z - 2| 
< 3<5 + 55 + 25 

= 105 < 10— = e. 
~ 10 

Thus, lim( x , VjZ )^( 1 ,_i i2 )(3x — 5y + 2z) = 12, as desired. ♦ 

Using the same methods as in Example 4, you can show that 

lim(fliJfi + (22*2 + ■ ■ ■ + a n x n ) = a\b\ + 02^2 + ■ ■ ■ + a„b n 

for any cij, i = 1,2, n. 

Some Topological Terminology 

Before discussing the geometric meaning of the limit of a function, we need to in- 
troduce some standard terminology regarding sets of points in R" . The underlying 
geometry of point sets of a space is known as the topology of that space. 

Recall from §2.1 that the vector equation ||x — a|| = r, where x and a are 
in R 3 and r > 0, defines a sphere of radius r centered at a. If we modify this 
equation so that it becomes the inequality 

Hx-a||<r, (1) 

then the points x e R 3 that satisfy it fill out what is called a closed ball shown in 
Figure 2.31. Similarly, the strict inequality 

||x-a||<r (2) 

describes points x e R 3 that are a distance of less than r from a. Such points 
determine an open ball of radius r centered at a, that is, a solid ball without the 
boundary sphere. 

There is nothing about the inequalities (1) and (2) that tie them to R 3 . In fact, 
if we take x and a to be points of R", then (1) and (2) define, respectively, closed 
and open n -dimensional balls of radius r centered at a. While we cannot draw 
sketches when n > 3, we can see what (1) and (2) mean when n is 1 or 2. (See 
Figures 2.32 and 2.33.) 



y y 




X 



Figure 2.32 The closed and open balls (disks) in R 2 defined by [|x — a[| < r and 



Chapter 2 | Differentiation in Several Variables 



X 



Figure 2.34 The graph of X. 









-s^ — 











Figure 2.37 The set X of 

Example 7. 



Figure 2.33 The closed and open balls (intervals) in R 
defined by \x — a\ < r and \x — a\ < r. 



DEFINITION 2.3 A set X C R" is said to be open in R" if, for each point 
x e X, there is some open ball centered at x that lies entirely within X. A 
point x e R" is said to be in the boundary of a set X c R" if every open ball 
centered at x, no matter how small, contains some points that are in X and 
also some points that are not in X. A set X cR" is said to be closed in R" 
if it contains all of its boundary points. Finally, a neighborhood of a point 
x g X is an open set containing x and contained in X. 



It is an easy consequence of Definition 2.3 that a set X is closed in R" precisely 
if its complement R" — X is open. 

EXAMPLE 5 The rectangular region 

X = {(x, y) g R 2 | -1 < x < 1, -1 < y < 2} 

is open in R 2 . (See Figure 2.34.) Each point in X has an open disk around it 
contained entirely in the rectangle. The boundary of X consists of the four sides 
of the rectangle. (See Figure 2.35.) ♦ 



X 



Figure 2.35 Every open disk 
about a point on a side of rectangle 
X of Example 5 contains points in 
both X and R 2 — X. 



Figure 2.36 The set X 

of Example 6 consists of 
the nonnegative 
coordinate axes. 



EXAMPLE 6 The set X consisting of the nonnegative coordinate axes in R 3 in 
Figure 2.36 is closed since the boundary of X is just X itself. ♦ 

EXAMPLE 7 Don't be fooled into thinking that sets are always either open or 
closed. (That is, a set is not a door.) The set 

X = {(x, y) g R 2 | 0 < x < 1, 0 < y < 1} 

shown in Figure 2.37 is neither open nor closed. It's not open since, for example, 
the point Q, 0) that lies along the bottom edge of X has no open disk around it 
that lies completely in X. Furthermore, X is not closed since the boundary of X 
includes points of the form (x, 1) for 0 < x < 1 (why?), which are not part of X. 



2.2 | Limits 103 



The Geometric Interpretation of a Limit 

Suppose that He R" — » R m . Then the geometric meaning of the statement 

lim f(x) = L 

\ Ml 

is as follows: Given any e > 0, you can find a corresponding S > 0 such that if 
points x e X are inside an open ball of radius 8 centered at a, then the correspond- 
ing points f(x) will remain inside an open ball of radius e centered at L. (See 
Figure 2.38.) 

y 





Figure 2.38 Definition of a limit: Given an open ball 5 e centered at L (right), you 
can always find a corresponding ball Bg centered at a (left), so that points in B s n X 
are mapped by f to points in B e . 

We remark that for this definition to make sense, the point a must be such 
that every neighborhood of it in R" contains points xeX distinct from a. Such 
a point a is called an accumulation point of X. (Technically, this assumption 
should also be made in Definition 2.2.) A point a e X is called an isolated point 
of X if it is not an accumulation point, that is, if there is some neighborhood of a 
in R" containing no points of X other than a. 

From these considerations, we see that the statement lim x ^ a f(x) = L really 
does mean that as x moves toward a, f(x) moves toward L. The significance of 
the "open ball" geometry is that entirely arbitrary motion is allowed. 

EXAMPLE 8 Let /: R 2 - {(0, 0)} -> R be defined by 



/(*. y) = 



y 



x 2 + y 2 

Let's see what happens to / as x = (x, y) approaches 0 = (0, 0). (Note that / is 
undefined at the origin, although this is of no consequence insofar as evaluating 
limits is concerned.) Along the x-axis (i.e., the line y = 0), we calculate the value 
of / to be 

,•2 



f(x, 0) = 



0 



0 



1. 



Thus, as x approaches 0 along the line y = 0, the values of / remain constant, 
and so 

lim /(x)=l. 

x— > 0 along y=0 

Along the j-axis, however, the value of / is 



m y) 



0-y 2 
0 + y 2 



1 04 Chapter 2 | Differentiation in Several Variables 



Hence, 



lim /(x) = -l. 

>0 along x=0 



Indeed, the value of / is constant along each line through the origin. Along the 
line y = mx, m constant, we have 



f(x, mx) 

Therefore, 



x 2 — m 2 x 2 x 2 (l — m 2 ) 1 — m 2 



x 2 +m 2 x 2 x 2 (\ + m 2 ) 1 + m 2 



1 - m 2 

lim f(x) 



x^O along y=mx 1 -|- m 

As a result, the limit of / as x approaches 0 does not exist, since / has different 
"limiting values" depending on which direction we approach the origin. (See 
Figure 2.39.) That is, no matter how close we come to the origin, we can find 
points x such that f(x) is not near any number LeR, (In other words, every 
open disk centered at (0, 0), no matter how small, is mapped onto the interval 
[ — 1 , 1].) If we graph the surface having equation 

2 2 

x — y 
x 2 + y 2 

(Figure 2.40), we can see quite clearly that there is no limiting value as x ap- 
proaches the origin. ♦ 



y 

R 2 




Figure 2.39 The function f{x, y) = (x 2 - y 2 )/(x 2 + y 2 ) of Figure 2.40 The graph of f(x, y) = 

Example 8 has value 1 along the jc-axis and value - 1 along the (* - y )/(* + y 1 ) of Example 8. 

y-axis (except at the origin). 



Warning Example 8 might lead you to think you can establish that 
lim^a f(x) = L by showing that the values of f as x approaches a along straight- 
line paths all tend toward the same value L. Although this is certainly good 
evidence that the limit should be L, it is by no means conclusive. See Exercise 23 
for an example that shows what can happen. 

EXAMPLE 9 Another way we might work with the function f(x, y) = (x 2 — 
y 2 )/(x 2 + y 2 ) of Example 8 is to rewrite it in terms of polar coordinates. Thus, 
let x = r cos 9, y = r sin 9. Using the Pythagorean identity and the double angle 



2.2 | Limits 105 

formula for cosine, we obtain, for r ^ 0, that 

x 2 - y 2 r 2 cos 2 9 - r 2 sin 2 9 r 2 (cos 2 9 - sin 2 9) cos 29 

= ~ = — cos 2$ 

x 2 + y 2 r 2 cos 2 (9 + r 2 sin 2 (9 r 2 (cos 2 6» + sin 2 6») 1 

That is, for r / 0, 

f(x,y) = f(r cos0, r sin#) = cos 20. 

Moreover, to evaluate the limit of / as (x, y) approaches (0, 0), we only must 
have r approach 0; there need be no restriction on 9. Therefore, we have 

lim f(x, y) = limcos2# = cos2#. 

This result clearly depends on 9. For example, if 9 = 0 (which defines the x-axis), 
then 

lim cos 29 = 1, 

r^O along 6 = 0 

while if 9 = tt/4 (which defines the line y = x), then 

lim cos 29 = 0. 

r^0 along 6 = jr/4 

Thus, as in Example 8, we see that lim (A V )^(o,o) f(x, y) fails to exist. ♦ 

EXAMPLE 10 We use polar coordinates to investigate lim( A>Y) ^(o,o) f(x, y), 
where f(x , y) = (x 3 + x 5 )/(x 2 + y 2 ). 

We first rewrite the expression (x 3 + x 5 )/(x 2 + y 2 ) using polar coordinates: 



x 3 + x 5 r 3 cos 3 0 + r 5 cos 5 1 



x 2 + y 2 r 2 cos 2 6> + r 2 sin 
Now — 1 < cos# < 1, which implies that 



r(cos 3 9 + r 2 cos 5 9). 



Hence, 



■ 1 - r 2 < cos 3 9 + r 2 cos 5 9 < 1 + r 2 . 



-r(\ + r 2 )< f(x,y)<r(l+r 2 ). 



As r -> 0, both the expressions — r(l + r 2 ) and r(l + r 2 ) approach zero. Hence, 
we conclude that lim (x , v )^ ( o,o) /(x, y) = 0, since / is squeezed between two 
expressions with the same limit. ♦ 

Properties of Limits 



One of the biggest drawbacks to Definition 2.2 is that it is not at all useful for 
determining the value of a limit. You must already have a "candidate limit" in 
mind and must also be prepared to confront some delicate work with inequalities 
to use Definition 2.2. The results that follow (which are proved in the addendum 



Chapter 2 | Differentiation in Several Variables 



to this section), plus a little faith, can be quite helpful for establishing limits, as 
the subsequent examples demonstrate. 



THEOREM 2.4 (Uniqueness OF limits) If a limit exists, it is unique. That is, 
let f: X C R" R m . If lim x ^ a f(x) = L and lim x ^ a f(x) = M, then L = M. 



THEOREM 2.5 (Algebraic properties) Let F,G:Xc R" -> R" be vector- 
valued functions, f,g:X^ R" -> R be scalar-valued functions, and let k € R 
be a scalar. 

1. If lim x ^ a F(x) = L and lim x ^ a G(x) = M, 
then lim x ^ a (F + G)(x) = L + M. 

2. If linw a F(x) = L, then lim x ^ a k¥(x) = kh. 

3. If lim x ^ a f(x) = L and lim x ^ a g(x) = M, then lim x ^ a (/g)(x) = LM. 

4. If lim x ^ a /(x) = L, g(x) / 0 for x e X, and lim x ^ a g(x) = M/ 0, then 
linw a (//g)(x) = L/M. 



There is nothing surprising about these theorems — they are exactly the same 
as the corresponding results for scalar- valued functions of a single variable. More- 
over, Theorem 2.5 renders the evaluation of many limits relatively straightforward. 

EXAMPLE 1 1 Either from rigorous considerations or blind faith, you should 
find it plausible that 

lim x = a and lim v = b. 

(x,y)^(a,b) (x,y)^(a,b) 

From these facts, it follows from Theorem 2.5 parts 1, 2, and 3 that 

lim (x 2 + 2xy - y 3 ) = a 2 + lab - b 3 , 

(x,y)-*(a,b) 

because, by part 1 of Theorem 2.5, 

lim (x 2 + 2xy - y 3 ) = limx 2 + lim2jcy + lim(-y 3 ) 

(x,y)-^(a,b) 

and, by parts 2 and 3, 

lim (x 2 + Ixy - y 3 ) = (limx) 2 + 2(limx)(limy) - (limy) 3 

(x,y)-*(a,b) 

so that, from the facts just cited, 

lim (x 2 + 2xy - y 3 ) = a 2 + lab - b 3 . ♦ 

(x, y y+(a,b) 

EXAMPLE 12 More generally, a polynomial in two variables x and y is any 
expression of the form 

d d 

p(x,y) = J2J2 c kix k y l , 

k=0 1=0 

where d is some nonnegative integer and c« 6 R for k, I = 0, . . . , d. That is, 
p(x, y) is an expression consisting of a (finite) sum of terms that are real number 
coefficients times powers of x and y. For instance, the expression x 2 + 2xy — y 3 
in Example 11 is a polynomial. For any (a, b) e R 2 , we have, by part 1 of 



2.2 | Limits 107 



Theorem 2.5, 

d d 

lim p(x, v) = y^y^ lim (ctix k y l ) 
so that, from part 2, 

d d 

lim p(x, y) = 7^ 7^ cm lim x k y l 

(x,y)->(a,b) f^ 0 (x,y)^(a,b) 

and, from part 3, 



d d 

lim p(x, j) = VV c H (limx*)(limy z ) 
= ±±c kl a k bK 

k=0 1=0 

Similarly, a polynomial in n variables x\, x%, . . . , x n is an expression of the form 



p(x\,x 2 , . . . ,x n ) = ^ c* :i ..4 n xf 1 x 2 2 • • •**" 



fei,...,ft„=0 

where is some nonnegative integer and c*,...^ 6 R for fci, . . . , fc„ = 0, . . . , d. 
For example, a polynomial in four variables might look like this: 

p{X\, . . . , Xa) = 1x\x-i + — 7X3X4. 

Theorem 2.5 implies readily that 

lim -k, X\ x 2 ■■■ x n " = Ck v -K a \ a 2 ' ' ' a „ ■ 

EXAMPLE 13 We evaluate lim * +Xy + ] . 

(x, y y+(-l,0) x 2 y - 5xy + y 2 + 1 

Using Example 12, we see that 

lim x 2 + xv + 3 = 4, 

(x,yH-(-l,0) 

and 

lim x 2 y - 5xy + y 2 + 1 = 1(/ 0). 

(A-,y)^<-1.0) ' 

Thus, from part 4 of Theorem 2.5, we conclude that 

lim * 2 + '? + 3 = 1 = 4 . 

(x,y)^(-i,0) x 2 y - 5xy + y 2 + 1 1 

EXAMPLE 14 Of course, not all limits of quotient expressions are as simple 
to evaluate as that of Example 13. For instance, we cannot use Theorem 2.5 to 
evaluate 

2 4 
x — y 

lim — j (3) 

since ]jm( Xt yy+(o,o)(x 2 + y 4 ) = 0. Indeed, since lim^^-^o.o^x 2 — y 4 ) = 0 as 
well, the expression (x 2 — y 4 )/(x 2 + y 4 ) becomes indeterminate as (x, y) -> 
(0, 0). To see what happens to the expression, we note that 

x 2 4 x 2 
lim — = lim — = 1, 

.1-^0 along y=0 X 2 + y 4 x-*0 X 2 



Chapter 2 | Differentiation in Several Variables 



while 



lim 

>0 along 



=o x 2 + y 4 



= lim 



v 4 



Thus, the limit in (3) does not exist. (Compare this with Example 8.) ♦ 

The following result shows that evaluating the limit of a function 
f : X c R" —> R" is equivalent to evaluating the limits of its (scalar- valued) com- 
ponent functions. First recall from §2.1 that f(x) may be rewritten as (/i(x), 
/ 2 (x), . . . , / m (x)). 

THEOREM 2.6 Suppose He R" R" 1 is a vector-valued function. Then 
lim x ^ a f(x) = L, where L = (Li, . . . , L m ), if and only if lim x ^ a f t (x) = Li for 
i = 1, .... m. 



EXAMPLE 15 Consider the linear mapping f: R" -■>• R" defined by f(x) = Ax, 
where A = (a, ; ) is an m x n matrix of real numbers. (See Example 5 of §1.6.) 
Theorem 2.6 shows us that 



for any h = (b\ 



lim f(x) = Ab 

x— >b 

, b n ) in R" . If we write out the matrix multiplication, we have 



f(x) = Ax = 



flu 

«21 
_ Ami 



flln 
fl2;i 



-Vl 

A'2 



mn _1 l_ -^n _ 



a U X\ + (312^2 + 
Cl2\X\ + a 2 2*2 + 



~f" Cl\n%n 



flml-^-1 ~T~ @m2X2 1***1 0. mn X n 

Therefore, the tth component function of f is 

= anxi + a,- 2 x 2 H h fli«x„. 

From Example 4, we have that 

lim yi(x) = fln^i + fl/2^2 H h a,>A 

x^b 

for each /. Hence, Theorem 2.6 tells us that the limits of the component functions 
fit together to form a limit vector. We can, therefore, conclude that 



limf(x) = (lim/ 1 (x), . 

x^b x^b 

= (aubi H 

a n b] + 
a 2 \b\ + 



, lim / m (x)) 

\ > b 

■ ci\ n b n , . . . , a m \b\ + 

+ a\ n b n 
+ a 2n b n 



®mnb\\) 



a m \b\ + • • ■ + a„,„b„ 



once we take advantage of matrix notation. 



Ab, 



2.2 | Limits 109 



Figure 2.41 The graph of a 
continuous function. 




Figure 2.44 The graph of a 
continuous function f(x, y). 



Continuous Functions 

For scalar-valued functions of a single variable, one often adopts the following 
attitude toward the notion of continuity: A function /: X C R is continuous 
if its graph can be drawn without taking the pen off the paper. By this criterion, 
Figure 2.41 describes a continuous function y = f(x), while Figure 2.42 does not. 






Figure 2.42 The graph of a 
function that is not continuous. 



Figure 2.43 The graph of / 
where f(x, y) = 0 if both x > 0 
and y > 0, and where f(x, y) = 1 
otherwise. 



We can try to extend this idea to scalar-valued functions of two variables: A 
function /: X C R 2 — R is continuous if its graph (in R 3 ) has no breaks in it. 
Then the function shown in Figure 2.43 fails to be continuous, but Figure 2.44 
depicts a continuous function. Although this graphical approach to continuity is 
pleasantly geometric and intuitive, it does have real and fatal flaws. For one thing, 
we can't visualize graphs of functions of more than two variables, so how will we 
be able to tell in general if a function /:XcR"^ R m is continuous? Moreover, 
it is not always so easy to produce a graph of a function of two variables that is 
sufficient to make a visual determination of continuity. This said, we now give a 
rigorous definition of continuity of functions of several variables. 



DEFINITION 2.7 Let f: X C R" -> R" ! and let a e X. Then f is said to be 
continuous at a if either a is an isolated point of X or if 

limf(x) = f(a). 

x->a 

If f is continuous at all points of its domain X, then we simply say that f is 
continuous. 



EXAMPLE 16 Consider the function /: R 2 

x 2 + xy — 2y 2 

f(x,y)= ' 



x 2 + y 2 



0 



-> R defined by 
if (x, y) # (0, 0) 
if (x, y) = (0, 0) 



Therefore, f(0, 0) = O^utlim^^^o) f(x, y) does not exist. (To see this, check 
what happens as (x, y) approaches (0,0) first along y = 0 and then along x = 0.) 
Hence, / is not continuous at (0,0). ♦ 

It is worth noting that Definition 2.7 is nothing more than the "vectorized" 
version of the usual definition of continuity of a (scalar-valued) function of one 
variable. This definition thus provides another example of the power of our vector 
notation: Continuity looks the same no matter what the context. 



Chapter 2 | Differentiation in Several Variables 



One way of thinking about continuous functions is that they are the ones whose 
limits are easy to evaluate: When f is continuous, the limit of f as x approaches 
a is just the value of f at a. It's all too tempting to get into the habit of behaving 
as if all functions are continuous, especially since the functions that will be of 
primary interest to us will be continuous. Try to avoid such an impulse. 

EXAMPLE 1 7 Polynomial functions in n variables are continuous. Example 12 
gives a sketch of the fact that 

lim y* c kv .. ki x\ 1 ■ ■ ■ xl" = y~] c h ... ka a^ ■ ■ ■ a*", 

where x = (xi, . . . , x n ) and a = (#i, . . . , a n ) are in R". If /: R" —> R is defined 
by 

/(x) = X>*,... fc *?'---#. 
then the preceding limit statement says precisely that / is continuous at a. ♦ 

EXAMPLE 18 Linear mappings are continuous. If f: R" — > R'" is denned by 
f(x) = Ax, where A is an m x n matrix, then Example 1 5 establishes that 

lim f(x) = Ab = f(b) 

for all b e R". Thus, f is continuous. ♦ 

The geometric interpretation of the e — S definition of a limit gives rise to a 
similar interpretation of continuity at a point: f: X C R" —> R" ! is continuous at 
a point a £ X if, for every open ball B € in R" of radius e centered at f(a), there 
is a corresponding open ball B s in R" of radius 8 centered at a such that points 
x e X inside B & are mapped by f to points inside B ( . (See Figure 2.45.) Roughly 
speaking, continuity of f means that "close" points in X c R" are mapped to 
"close" points inR ffl . 




In practice, we usually establish continuity of a function through the use 
of Theorems 2.5 and 2.6. These theorems, when interpreted in the context of 



2.2 | Limits 1 1 1 



continuity, tell us the following: 



• The sum F + G of two functions F, G: X C R" — > R m that are continuous 
at a e X is continuous at a. 

• For all & € R, the scalar multiple kF of a function F:XcR"->R ffl that 
is continuous at a £ X is continuous at a. 

• The product fg and the quotient f/g(gj^ 0) of two scalar- valued functions 
f,g:XC. R" R that are continuous ata e I are continuous at a. 

• F:XcR"-> R'" is continuous at a e X if and only if its component 
functions F* : X C R" — »• R, j = 1 , . . . , m are all continuous at a. 



EXAMPLE 1 9 The function f: R 2 -> R 3 denned by 

f(x, y) = (x + y, x 2 y, y sin(xy)) 

is continuous. In view of the remarks above, we can see this by checking that the 
three component functions 

fi(x,y) = x + y, f 2 (x,y) = x 2 y, and f 3 (x, y) = y sin(xy) 

are each continuous (as scalar-valued functions). Now f\ and fi are continuous, 
since they are polynomials in the two variables x and y. (See Example 17.) The 
function is the product of two further functions; that is, 

Mx, y) = g(x, y)h(x, y), 

where g{x, y) = y and h(x, y) = sin(xy). The function g is clearly continuous. 
(It's a polynomial in two variables — one variable doesn't appear explicitly!) The 
function h is a composite of the sine function (which is continuous as a function 
of one variable) and the continuous function p(x, y) = xy. From these remarks, 
it's not difficult to see that 

lim h(x, y) = lim sin(p(x, y)) 

(x,y)-+(a,b) (x,y)^(a,b) 

= sin ( lim p(x, y) I , 

since the sine function is continuous. Thus, 

lim h(x, y) = sin p(a, b) = h(a, b), 

because p is continuous. Thus, h, hence fa, and, consequently, fare all continuous 
on all of R 2 . ♦ 

The discussion in Example 19 leads us to the following general result, whose 
proof we omit: 



THEOREM 2.8 Iff: XcR"^ R"< andg: Y C R" RP are continuous func- 
tions such that range fey, then the composite function g o f: X C R" — > R p is 
defined and is also continuous. 



Chapter 2 | Differentiation in Several Variables 



Addendum: Proofs of Theorems 2.4, 2.5, 2.6, and 2.8 

For the interested reader, we establish the various results regarding limits of 
functions that we used earlier in this section. 

Proof of Theorem 2.4 The statement lim x ^ a f(x) = L means that, given any e > 
0, we can find some Si > 0 such that if x 6 X and 0 < ||x — a|| < Si, then ||f(x) — 
L|| < e/2. (The reason for writing e/2 rather than e will become clear in a 
moment.) Similarly, lim x ^ a f(x) = M means that, given any e > 0, we can find 
some S 2 > 0 such that if x e X and 0 < ||x - a|| < S 2 , then ||f(x) - M|| < e/2. 

Now let S = min^!, £2); that is, we set S to be the smaller of Si and S 2 . If 
x e X and 0 < ||x - a|| < 8, then both ||f(x) - L|| and ||f(x) - M|| are less than 
e /2 so that, using the triangle inequality, we have 

||L-M|| = ||(L-f(x)) + (f(x)-M)|| 

< ||L-f(x)|| + ||f(x)-M|| <| + | = 6 . 

This shows that the quantity ||L — M|| can be made arbitrarily small; thus, it 
follows that L - M = 0. Hence, L = M. ■ 

Proof of Theorem 2.5 To establish part 1, note that if lim x ^ a F(x) = L, then 
given any e > 0, we can find a Si > 0 such that if x e X and 0 < ||x — a|| < 
Si, then || F(x) — L|| < e/2. Similarly, if lim x ^ a G(x) = M, then we can find a 
S 2 > 0 such that if x 6 X and 0 < ||x - a|| < S 2 , then ||G(x) - M|| < e/2. Now 
let i5 = min((5i, ^2). Then if x e X and 0 < ||x — a|| < <5, the triangle inequality 
implies that 

||(F(x) + G(x)) - (L + M)|| < ||F(x) - L|| + ||G(x) - M\\ <"-+"-= e. 

Hence, lim x ^ a (F(x) + G(x)) = L + M. 

To prove part 2, suppose that e > 0 is given. If lim x ^ a F(x) = L, then we can 
find a <5 > 0 such that if x £ X and 0 < ||x - a|| < S, then ||F(x) - L|| < e/\k\. 
Therefore, 

\\k¥(x) - *L|| = |*| ||F(x) - L|| < |^| -1 = e, 

\k\ 

which means that lim x ^ a k¥(x) = kL,. (Note: If k = 0, then part 2 holds trivially.) 

To establish the rule for the limit of a product of scalar- valued functions (part 
3), we will use the following algebraic identity: 

/(x)g(x) -LM = (/(x) - L)(g(x) — M) + L(g(x) — M) + M(f(x) - L). 

(4) 

If lim x ^ a f(x) = L, then, given any e > 0, we can find Si > 0 such that if x e X 
and 0 < ||x — a|| < Si, then 

|/(x)-L| < Vi\ 

Similarly, if lim x ^ a g(x) = M, we can find S 2 > 0 such that if x e X and 0 < 
||x — a|| < S 2 , then 

\g(x)-M\ < Je. 
Let 8 = min((5i , ^2). If x e X and 0 < ||x — a|| < 8, then 



|(/(x) - L)(g(x) -M)\ <s Te-^e = e. 



2.2 | Limits 1 1 3 



This means that lim x ^ a (/(x) — L)(g(x) — M) = 0. Therefore, using (4) and parts 
1 and 2, we see that 

lim(/(x)g(x) - LM) = lim(/(x) - L)(g(x) -M) + L lim(g(x) - M) 

x^- a x-> a x-^- a 

+ M lim(/(x) - L) 

x— >a 

= 0 + 0 + 0 = 0. 

Since lim x ^ a f(x)g(x) = lim x ^ a ((/(x)^(x) — LM) + LM), the desired result 
follows from part 1 . 

The crux of the proof of part 4 is to show that 



lim 



1 



1 

~M 



x^a g(x) 

Once we show this, the desired result follows directly from part 3 : 



/(x) 



lim 

x^a g( X ) 



Note that 



= lim /(X) 



1 

M 



1 



S(x) 



S(x) 

_ \M - g(x)\ 
\Mg(x)\ 



1 _ L 

~ ~M ~ ~M 



and, by the triangle inequality, that 

\M\ = \M- g(x) + g(x)| <\M- g(x)\ + \g(x) 



(5) 



If lim x ^. a g(x) = M, then, given any e > 0, we can find 8\ such that if x e X and 
0 < ||x- a|| < <5!,then 

M 2 

\g(x)-M\ <—e. 

We can also fmd 82 such that if x e X and 0 < ||x — a|| < 82, then |g(x) — M\ < 
\M\/2 and, hence, using (5), that 



|M| 

\M\ < + 1*001 \g(x)\ > 



\M\ 



1 



Now let 8 = min(<5i , ^2). If x e X and 0 



1 



g(x) 



1 

M 



|M-g(x)| 
|Mg(x)| 
1 2 M 2 



2 l#(x)| 
x — a|| < 8, then 

1 |M-g(x)| 
" IM| 



< 



|M| 



lg(x)l 



< |M| |M| 2 e € ' 



Proof of Theorem 2.6 Note first that, for i = 1, . . . , m, 

\fi(x) ~L,\< V(/ 1 (x)-L 1 ) 2 + --- + (/ m (x)-L m ) 2 = ||f(x) - L|| 



(6) 



If lim x ^ a f(x) = L, then given any e > 0, we can find a 5 > 0 such that if x e 
X and 0 < ||x - a|| < 8, then ||f(x) - L|| < e. Hence, (6) implies that |/;(x) - 
Lf\ < € for i = 1, . . . , m, which means that lim x ^ a /,(x) = L ; . 

Conversely, suppose that lim x ^ a /i(x) = L,- for i = 1, . . . , m. This means 
that, given any e > 0, we can find, for each i , a 8 t > 0 such that if x e X and 



14 Chapter 2 | Differentiation in Several Variables 



0 < ||x — a|| < 8i, then |/}(x) — L,| < e/^Jm. Set 8 = min(<5i, . . . , 8 m ). Then if 
x e X and 0 < ||x — a|| < 8, we see that (6) implies 

||f(x)-L|| < J- + = m - = € . 

V m m V m 

Thus, lim x ^ a f(x) = L. ■ 

Proof of Theorem 2.8 We must show that the composite function g o f is con- 
tinuous at every point a e X. If a is an isolated point of X, there is nothing to 
show. Otherwise, we must show that lim x ^ a (g o f)(x) = (go f)(a). 

Given any e > 0, continuity of g at f(a) implies that we can find some y > 0 
such that if y e range f and 0 < ||y — f(a)|| < y then 

||g(y)-g(f(a))|| <6. 

Since f is continuous at a, we can find some <5 > 0 such that if x e X and 0 < 
|| x — a|| < 8, then 

||f(x)-f(a)|| < y. 
Therefore, if x e X and 0 < ||x — a|| < 8, then 

||g(f(x))-g(f(a))|| <e. m 

2.2 Exercises 



In Exercises 1—6, determine whether the given set is open or 
closed (or neither). 

1. {(x,y)eR 2 | 1 < x 2 + y 2 < 4} 

2. {(x,y)eR 2 | 1 <x 2 +y 2 < 4} 

3. {(x,y)eR 2 | 1 < x 2 + y 2 < 4} 

4. {(x, y, z) e R 3 | 1 < x 2 + y 2 + z 2 < 4} 

5. {(x, y) e R 2 | -1 < x < 1} U {(x, y) e R 2 | 

6. {(x,y,z)eR 3 | 1 < x 2 + y 2 < 4} 



2} 



Evaluate the limits in Exercises 7—21, or explain why the limit 
fails to exist. 

x 2 + 2xy + yz + z 3 + 2 



7. lim 

(*,.y,z)-H0,0,0) 

8. lim 



l.v| 



(x,y)->(0,0) j x l _|_ yl 

9. lim — - 

(x.y)^r(P.O) x 2 + y 2 

10. lim 



(jt.yH-ro.O) x + y + 2 
2x 2 + y 2 



1 1 . lim 



(jt,y)-+(0.0) X 2 + y 2 

2x 2 + y 2 



12. lim 



(x,y)^(-l,2) X 2 + y 2 



x 2 + 2xy + y 2 
u-O -to ui x + y 

xy 



13. lim 

14. lim 



(*,;>>)-K0,0) x 2 + y 2 

x 4 -y 4 

15. lim — ■— 

(jr,y)-<-(0,0) x 1 + y 1 
X 2 

16. lim — - 

(x,y)^(0,0) X 2 + y 2 



17. 



lim 



x — xy 



(jc,y)->(o.o) > *54> -Jx - 

x 2 -y 2 -4x+4 
1 8. lim — — 

(x,y)^(2,0) x 2 +y 2 -4x +4 



19. 

20. 
21. 



lim e xz cos y — x 

(x,y,z)^(0,^fit.l) 



2x 2 + 3y 2 + z 2 

(jc,y,t)->-(0,0,a) X 2 + y 2 + Z 2 
xy — xz + yz 



lim 



lim 



(jc,y,t)-K0,0,a) x 2 + y 2 + z 2 
sin© 

22. (a) What is lim ? 

(b) What is lim Sm(x + y) ? 

(x,y)^(0,0) x + y 

(c) What is lim 

(x,y)^(0,0) xy 



2.2 | Exercises 115 



23. Examine the behavior of fix, y) = x 4 y 4 /(x 2 + y 4 ) 3 
as (x, y) approaches (0, 0) along various straight lines. 
From your observations, what might you conjecture 
lim( XJ ,)_i.(o i o) f(x, y) to be? Next, consider what hap- 
pens when (x, y) approaches (0, 0) along the curve 
x = y 2 . Does lim^- yj^^o) fix, y) exist? Why or why 
not?' 

In Exercises 24—27, (a) use a computer to graph z = f(x, y); 
(b) use your graph in part (a) to give a geometric discussion 
as to whether lim^ ^^^o) f(x, y) exists; (c) give an analytic 
(i.e., nongraphical) argument for your answer in part (b). 

4x 2 + 2xy + 5 y 2 



^ 24. f(x, y) : 

^25. f(x,y)-. 

^ 26. f{x, y) : 

27. f(x, y) : 



3x 2 + 5y 2 



y 



x 2 + y 2 

„5 



xy 



x 2 + v 1 



x sin — if y ^ 0 



y 



o 



ify = 0 



Some limits become easier to identify if we switch to a different 
coordinate system. In Exercises 28-33 switch from Cartesian to 
polar coordinates to evaluate the given limits. In Exercises 34— 
37 switch to spherical coordinates. 



28. 



29. 



30. 



31. 



32. 



33. 



34. 



35. 



36. 



37. 



lim 



lim 

*,v>^(0.0) 



lim 



lim — - 

z,}>)->-(0,0) x 2 + y 2 

x 2 

lim — r 

x,y)-y(0,0) X 2 + y 2 



x 2 + xy + y 2 
x 2 +y 2 

x 5 + y 4 - 3x 3 y + 2x 2 + 2y 2 
x 2 + y 2 



x + y 

lim 



lim 



2 

x y 



x,j, z )^(0,o,o) x 2 + y 2 + z 2 
xyz 

lim — 

x,y,z)^(0,0,0) x 2 + y 2 + z 2 



lim 



x 2 + y 2 



*,y,z)-K0,0.0) Jx 2 + y 2 +z 2 

XZ 

lim — = - 

>,y,z)^-(0,0,0) X 2 + V 2 + Z 2 



In Exercises 38—45, determine whether the functions are con- 
tinuous throughout their domains: 

38. fix,y) = x 2 + 2xy - y 1 

39. fix, y, z) = x 2 + 3xyz + yz 3 +2 



40. g(x,y): 



x 2 + \ 



41. h(x, y) = cos 



x 2 -y 2 
x 2 + \ 



42. fix, y) = cos 2 x — 2 sin 2 xy 

if(x,y)/(0,0) 
0 if(x,y) = (0,0) 
x 3 + x 2 + xy 2 + y 



43. fix, y) : 



x 2 -y 2 
x^Ty 2 



44. g(x,y) . 



x 2 + y 2 
2 



45. F(x,y,z)= [x z + 3xy 



if(x,y)/(0, 0) 
if(x,y) = (0, 0) 

xy 



2x 2 + y 4 + 3 ' \v 2 + l 

46. Determine the value of the constant c so that 

' x 3 +xy 2 +2x 2 +2y 2 



g(x, y) = 



is continuous. 



x + y 2 
c 



if ix, y) # (0, 0) 
if ix, y) = (0, 0) 



47. Show that the function /: R 3 — > R given by f(\) = 
(2i — 3j + k) • x is continuous. 

48. Show that the function f: R 3 — > R 3 given by f(x) = 
(6i — 5k) x x is continuous. 

Exercises 49—53 involve Definition 2.2 of the limit. 

49. Consider the function fix) = 2x — 3. 

(a) Show that if \x - 5| < 5, then |/(jc) - 7| < 25. 

(b) Use part (a) to prove that lim A -^ 5 fix) = 7. 

50. Consider the function fix, y) = 2x — lOy + 3. 

(a) Show that if ||(jc, y) - (5, 1)|| < S, then \x - 5| < 
S and |y — 1 1 < S. 

(b) Use part (a) to show that if ||(jc, y) - (5, 1)|| < S, 
then|/(A,y)- 3| < 125. 

(c) Show that lim (x } ,)^(5i) fix, y) = 3. 

51 . If A, B, and C are constants and fix, y) = Ax + By + 
C, show that 

lim fix, y) = fix 0 , y 0 ) = Ax 0 + By 0 + C. 

(jc,y)-s-(x 0 ,>'o) 



Chapter 2 | Differentiation in Several Variables 



52. In this problem, you will establish rigorously that 

lim — = 0. 

(x,y)M0,0) x 2 + y 2 

(a) Show that \x\ < \\{x,y)\\ and \y\ < \\(x,y)\\. 

(b) Showthat |jc 3 + y 3 \ < 2(x 2 + y 2 fl 2 . (Hint: Begin 
with the triangle inequality, and then use part (a).) 

(c) Show that if 0 < ||(jc, y)|| < S, then \(x 3 + y 3 )/ 
(x 2 + y 2 )\ < 28. 

(d) Now prove that \im (x v) ^ (0 o)(-* 3 + v 3 )/(x 2 + 

y 1 ) = o. 



53. (a) If a and b are any real numbers, show that 2\ab\ < 
a 2 + b 2 . 

(b) Let 

(x 2 - y 2 \ 

Use part (a) to show that if 0 < ||(jc, y)\\ < S, then 
\f(x,y)\ <S 2 /2. 

(c) Prove that lim (A: ,_ V )_j.(o,o) f(x, y) exists, and find its 
value. 



2.3 The Derivative 

Our goal for this section is to define the derivative of a function f : X C R" — > R" , 
where n and m are arbitrary positive integers. Predictably, the derivative of a 
vector-valued function of several variables is a more complicated object than the 
derivative of a scalar- valued function of a single variable. In addition, the notion of 
differentiability is quite subtle in the case of a function of more than one variable. 

We first define the basic computational tool of partial derivatives. After do- 
ing so, we can begin to understand differentiability via the geometry of tangent 
planes to surfaces. Finally, we generalize these relatively concrete ideas to higher 
dimensions. 

Partial Derivatives 

Recall that ifF:XcR^Risa scalar- valued function of one variable, then the 
derivative of F at a number a e X is 

F(a + h)-F(a) 
F(a)=hm . (1) 

fc->o h 

Moreover, F is said to be differentiable at a precisely when the limit in equation 
(1) exists. 



DEFINITION 3.1 Suppose f:X C R" R is a scalar- valued function of n 
variables. Let x = (x\ , x%, . . . , x n ) denote a point of R" . A partial function 
F with respect to the variable Xj is a one-variable function obtained from 
/ by holding all variables constant except x, . That is, we set xj equal to a 
constant aj for j ^ i. Then the partial function in x, is defined by 

F (*i) = ,a 2 ,...,Xi,..., a n ). 



EXAMPLE 1 If/(x, y) = (x 2 - y 2 )/(x 2 + y 2 ), then the partial functions with 
respect to x are given by 



2 2 

X — ai 



F(x) = f(x,a 2 ) = 

x 1 + aj 



where a 2 may be any constant. If, for example, a 2 = 0, then the partial function is 

x 2 

F{x) = f(x, 0) = ^ = 1. 



2.3 | The Derivative 117 



y 



Domain of/ 




/ 

Domain of F 

(restriction 

of/) 





Figure 2.46 The function / of 
Example 1 is defined on R 2 - 
{(0,0)}, while its partial function F 
along y = 0 is defined on the 
jc-axis minus the origin. 




x 

Figure 2.47 Visualizing the 
partial derivative f^(a, b). 



z 




X 

Figure 2.48 Visualizing the 
partial derivative §r(a, b). 



Geometrically, this partial function is nothing more than the restriction of / to 
the horizontal line y = 0. Note that since the origin is not in the domain of /, 0 
should not be taken to be in the domain of F. (See Figure 2.46.) ♦ 

Remark In practice, we usually do not go to the notational trouble of explic- 
itly replacing the xj 's (j 7^ i) by constants when working with partial functions. 
Instead, we make a mental note that the partial function is obtained by allowing 
only one variable to vary, while all the other variables are held fixed. 



DEFINITION 3.2 The partial derivative of / with respect to x; is the 

(ordinary) derivative of the partial function with respect to x, . That is, the 
partial derivative with respect to x, is F'(x ; ), in the notation of Definition 
3.1. Standard notations for the partial derivative of / with respect to x,- are 

df 

~z > D Xi /(xi ,...,*„), and f Xi (x u x„). 
dx, 

Symbolically, we have 

9/ ,. f(xi,...,x i + h,...,x n )-f(xi,...,x n ) 

— = hm . (2) 

dxi h^o h 



By definition, the partial derivative is the (instantaneous) rate of change of / 
when all variables, except the specified one, are held fixed. In the case where / 
is a (scalar-valued) function of two variables, we can understand 

ox 

geometrically as the slope at the point (a, b, f(a, b)) of the curve obtained by 
intersecting the surface z = fix, y) with the plane y = b, as shown in Figure 2.47. 
Similarly, 

-^-(a, b) 
dy 

is the slope at (a, b, f(a, fr))ofthe curve formed by the intersection of z = f(x, y) 
and x = a, shown in Figure 2.48. 

EXAMPLE 2 For the most part, partial derivatives are quite easy to compute, 
once you become adept at treating variables like constants. If 

f(x, y) = x 2 y + cos(x + y), 

then we have 

— = 2xy - sin(x + y). 
dx 

(Imagine y to be a constant throughout the differentiation process.) Also 

df 7 

— = x — sin(x + y). 

dy 



Chapter 2 | Differentiation in Several Variables 



Figure 2.49 The tangent line to 
y = F(x) at x = a has equation 
y = F(a) + F'(a)(x - a). 



(Imagine x to be a constant.) Similarly, if g(x, y) = xy/(x 2 
quotient rule of ordinary calculus, we have 

(x 2 + y 2 )y - xy(2x) y(y 2 - 
g x (x, y) = - 



y 2 ), then, from the 



x 2 ) 



and 



?y(*> y ) = 



(x 2 + y 2 ) 2 
(x 2 + y 2 )x — xy(2y) 



(x 2 + y 2 ) 2 ' 
x(x 2 — y 2 ) 



(x 2 + y 2 ) 2 (x 2 + y 2 ) 2 ' 

Note that, of course, neither g nor its partial derivatives are defined at (0, 0). ♦ 

EXAMPLE 3 Occasionally, it is necessary to appeal explicitly to limits to eval- 
uate partial derivatives. Suppose /: R 2 —> R is defined by 



/(*. y) 



3x y — y 



x 2 + y 2 



if (x, y) # (0, 0) 
if(x,y) = (0,0) 



Then, for (x, y) ^ (0, 0), we have 
9/ 8xy 3 



3x (x 2 + y 2 ) 



2\2 



and 



df 
dy 



3x — 6x y 



2, ,2 



(x 2 + y 2 ) 2 



3/ 3/ 

But what should — (0, 0) and — (0, 0) be? To find out, we return to Definition 

dx dy 

3.2 of the partial derivatives: 



and 



df 
dy 



df 
dx 



(0, 0) 



(0,0) 



lim 



/(0 + fc.O) 



= hm 

h-+o h 



lim 

h^0 



f(0, 0 + h)- /(0, 0) 



lim 



lim -1 = 



Tangency and Differentiability 

If F: X c R ^ R is a scalar- valued function of one variable, then to have F 
differentiable at a number a e X means precisely that the graph of the curve 
y = F(x) has a tangent line at the point (a, F(a)). (See Figure 2.49.) Moreover, 
this tangent line is given by the equation 



y = F(a) + F'(a)(x - a). 



(3) 



If we define the function H(x) to be F(a) + F'(a)(x — a) (i.e., H(x) is the right 
side of equation (3) that gives the equation for the tangent line), then H has two 
properties: 

1. H(a) = F(a) 

2. H'(a) = F'(a). 

In other words, the line defined by y = H(x) passes through the point (a, F(a)) 
and has the same slope at (a, F(a)) as the curve defined by y = F(x). (Hence, 
the term "tangent line.") 

Now suppose /:lcR 2 ^Risa scalar-valued function of two variables, 
where X is open in R 2 . Then the graph of / is a surface. What should the tangent 
plane to the graph of z = f(x, y) at the point (a, b, f(a, b)) be? Geometrically, 



2.3 | The Derivative 119 



(a,b,f{a,b)) 




Figure 2.50 The plane tangent 
toz = f(x, y) at 
(a,b,f(a,b)). 



y = b 




Figure 2.51 The tangent plane at 
(a, b, f(a, b)) contains the lines 
tangent to the curves formed by 
intersecting the surface 
z = f(x, y) by the planes x = a 
and y = b. 



the situation is as depicted in Figure 2.50. From our earlier observations, we 
know that the partial derivative f x (a, b) is the slope of the line tangent at the 
point (a, b, f(a, b)) to the curve obtained by intersecting the surface z = f(x, y) 
with the plane y = b. (See Figure 2.51.) This means that if we travel along this 
tangent line, then for every unit change in the positive x -direction, there's a change 
of f x (a , b) units in the z-direction. Hence, by using formula ( 1) of § 1 .2, the tangent 
line is given in vector parametric form as 

li(0 = (a, b, /(a, b)) + 0, f x (a, b)). 
Thus, a vector parallel to this tangent line is 

u = i + f x (a,b)k. 

Similarly, the partial derivative f y (a, b) is the slope of the line tangent at the point 
(a, b, f(a, b)) to the curve obtained by intersecting the surface z = f(x, y) with 
the plane x — a. (Again see Figure 2.51.) Consequently, the tangent line is given 
by 

1 2 (0 = (a, b, f(a, b)) + t(0, 1, f y (a, b)), 
so a vector parallel to this tangent line is 

v = j + f y (a,b)k. 

Both of the aforementioned tangent lines must be contained in the plane tangent 
to z = f{x, y) at (a, b, f(a, b)), if one exists. Hence, a vector n normal to the 
tangent plane must be perpendicular to both u and v. Therefore, we may take n 
to be 

n = u x v = -f x (a, b)\- f y (a, b)\ + k. 

Now, use equation ( 1 ) of § 1 .5 to find that the equation for the tangent plane — that 
is, the plane through (a, b, f(a, b)) with normal n — is 

(-f x (a, b), -f y (a, b), 1) • (x - a, y - b, z - f(a, bj) = 0 

or, equivalently, 

-f x (a, b)(x -a)- f y (a, b)(y -b) + z- f(a, b) = 0. 

By rewriting this last equation, we have shown the following result: 



THEOREM 3.3 If the graph of z = f(x, y) has a tangent plane at 
(a, b, f(a, b)), then that tangent plane has equation 



z = f(a, b) + f x (a, b)(x - a) + f y (a, b)(y - b). 



(4) 



Note that if we define the function h(x, y) to be equal to f(a, b) + f x (a, b)(x — 
a) + f y (a, b)(y — b) (i.e., h(x, y) is the right side of equation (4)), then h has the 
following properties: 



1. h(a,b) = f(a,b) 
dh df 
ox ax 



dh df 
and — (a,b)= — (a, b). 
dy dy 



In other words, h and its partial derivatives agree with those of / at (a, b). 

It is tempting to think that the surface z = fix, y) has a tangent plane at 
(a, b, f(a, b)) as long as you can make sense of equation (4), that is, as long as the 



1 20 Chapter 2 | Differentiation in Several Variables 




Figure 2.52 If two points 
approach (0, 0, 0) while remaining 
on one face of the surface 
described in Example 4, the 
limiting plane they and (0, 0, 0) 
determine is different from the one 
determined by letting the two 
points approach (0, 0, 0) while 
remaining on another face. 



partial derivatives f x (a, b) and f y (a, b) exist. Indeed this would be analogous to 
the one-variable situation where the existence of the derivative and the existence 
of the tangent line mean exactly the same thing. However, it is possible for a 
function of two variables to have well-defined partial derivatives (so that equation 
(4) makes sense) yet not have a tangent plane. 

EXAMPLE 4 Let f(x, y) = \\x\ - \y\\ - \x\ - |y| and consider the surface 
defined by the graph of z = f(x, y) shown in Figure 2.52. The partial derivatives 
of / at the origin may be calculated from Definition 3.2 as 



A(0, 0) = lim 

h— >0 



/(0 + ft,0)-/(0, 0) 



= lim 



and 



f y (0, 0) = lim 

h^0 



/(O, 0 + h)- /(O, 0) 



= lim 

h^0 



h 



= lim 0 = 0 



= lim 0 = 0. 

h^0 



(Indeed the partial functions F(x) = f(x, 0) and G(y) = f(0, y) are both identi- 
cally zero and thus, have zero derivatives.) Consequently, if 'the surface in question 
has a tangent plane at the origin, then equation (4) tells us that it has equation 
z = 0. But there is no geometric sense in which the surface z = f(x, y) has a 
tangent plane at the origin. If we think of a tangent plane as the geometric limit of 
planes that pass through the point of tangency and two other "moving" points on 
the surface as those two points approach the point of tangency, then Figure 2.52 
shows that there is no uniquely determined limiting plane. ♦ 



Example 4 shows that the existence of a tangent plane to the graph of 
z = f(x, y) is a stronger condition than the existence of partial derivatives. It 
turns out that such a stronger condition is more useful in that theorems from the 
calculus of functions of a single variable carry over to the context of functions 
of several variables. What we must do now is find a suitable analytic definition of 
differentiability that captures this idea. We begin by looking at the definition of 
the one-variable derivative with fresh eyes. 

By replacing the quantity a + h by the variable x , the limit equation in formula 
(1) may be rewritten as 

F(x) - F(a) 
F (a) = hm . 



This is equivalent to the equation 

F(x) 



lim 



F(a) 



F'(a) = 0. 



The quantity F'(a) does not depend on x and therefore may be brought inside the 
limit. We thus obtain the equation 

F(x) - F(a) 



lim 



F'(a) =0. 



Finally, some easy algebra enables us to conclude that the function F is differen- 
tiable at a if there is a number F'(a) such that 



lim 



F(x) - [F(a) + F'(a)(x - a)] 



= 0. 



(5) 



What have we learned from writing equation (5)? Note that the expression in 
brackets in the numerator of the limit expression in equation (5) is the function 



2.3 | The Derivative 121 





(x,F(x)) 




^\(a, F(a)) \ 






(x, H(x)^ 






a x 



Figure 2.53 If F is differentiable 
at a, the vertical distance between 
F(x) and H(x) must approach 
zero faster than the horizontal 
distance between x and a does. 



H(x) that was used to define the tangent line to y 
may rewrite equation (5) as 



F(x) at (a, F(a)). Thus, we 



lim 



F{x) - H(x) 



= 0. 



For the limit above to be zero, we certainly must have that the limit of the numerator 
is zero. But since the limit of the denominator is also zero, we can say even more, 
namely, that the difference between the y-values of the graph of F and of its tangent 
line must approach zero faster than x approaches a. This is what is meant when 
we say that "H is a good linear approximation to F near a." (See Figure 2.53.) 
Geometrically, it means that, near the point of tangency, the graph of y = F(x) 
is approximately straight like the graph of y = H(x). 

If we now pass to the case of a scalar- valued function fix , y) of two variables, 
then to say that z = fix, y) has a tangent plane at (a, b, f(a, b)) (i.e., that / is 
differentiable at ia, b)) should mean that the vertical distance between the graph 
of / and the "candidate" tangent plane given by 

z = h(x, y) = fia, b) + f x (a, b){x -a)+ f y (a, b\y - b) 

must approach zero faster than the point (jc, y) approaches (a,b). (See Fig- 
ure 2.54.) In other words, near the point of tangency, the graph of z = fix, y) is 
approximately flat just like the graph of z = h(x, y). We can capture this geometric 
idea with the following formal definition of differentiability: 



DEFINITION 3.4 Let X be open in R 2 and /: X C R 2 -> R be a scalar- 
valued function of two variables. We say that / is differentiable at (a, b) e X 
if the partial derivatives f x (a, b) and f y (a, b) exist and if the function 



h( x < y) = f( a < b ) + fx(fl, b\x -a) + f y (a, b)(y 
is a good linear approximation to / near (a, b) — that is, if 

fix,y)-h(x,y) 



b) 



lim 



ix,y)-ia,b)\\ 



= 0. 



Moreover, if / is differentiable at (a, b), then the equation z = h(x, y) de- 
fines the tangent plane to the graph of / at the point (a, b, fia, b)). If / 
is differentiable at all points of its domain, then we simply say that / is 
differentiable. 




(x,y,f(x,y)) 

(x,y,h(x,y)) 
ia,b,fia,b)) 



Figure 2.54 If / is differentiable at (a, b), the distance 
between fix, y) and h(x, y) must approach zero faster than 
the distance between (x, y) and (a, b) does. 



Chapter 2 | Differentiation in Several Variables 



EXAMPLE 5 Let us return to the function f(x, y) = \\x\ - \y\\ - \x\ - \y\ of 
Example 4. We already know that the partial derivatives f x (0, 0) and / v (0, 0) 
exist and equal zero. Thus, the function h of Definition 3.4 is the zero function. 
Consequently, / will be differentiable at (0,0) just in case 

f(x,y)-h(x,y) f(x,y) 

hm = hm 

(a-,y)^(0,0) ||(*, y) - (0, 0)|| C*,y)->-(0,0) ||(x, y)|| 

\x\-\y\\-\x\-\y\ 



= lim 



(x,v)^(o,o) y x 2 + 



r 



is zero. However, it is not hard to see that the limit in question fails to exist. Along 
the line v = 0, we have 

f(x,y) IM-01- W-I0I 0 Q 



ll(*,y)ll \x 
but along the line y = x, we have 

f(x,y) = lk|-|x||-|x|-|x| = -2|*| = _^ 
\\(x,y)\\ s/x 2 + x 2 </2\x\ 

Hence, / fails to be differentiable at (0, 0) and has no tangent plane at (0, 0, 0)> 

The limit condition in Definition 3.4 can be difficult to apply in practice. 
Fortunately, the following result, which we will not prove, simplifies matters in 
many instances. Recall from Definition 2.3 that the phrase "a neighborhood of 
a point P in a set X" just means an open set containing P and contained in X. 

THEOREM 3.5 Suppose X is open in R 2 . If /: X —> R has continuous partial 
derivatives in a neighborhood of (a, b) in X, then / is differentiable at (a, b). 

A proof of a more general result (Theorem 3. 10) is provided in the addendum 
to this section. 

EXAMPLE 6 Let /(*, y) = x 2 + 2y 2 . Then df/dx = 2x and df/dy = Ay, 
both of which are continuous functions on all of R 2 . Thus, Theorem 3.5 implies 
that / is differentiable everywhere. The surface z = x 2 + 2y 2 must therefore 
have a tangent plane at every point. At the point (2, — 1), for example, this tangent 
plane is given by the equation 

z = 6 + 4(x - 2) - 4(y + 1) 

(or, equivalently, by Ax — Ay — z = 6). ♦ 

While we're on the subject of continuity and differentiability, the next result is 
the multivariable analogue of a familiar theorem about functions of one variable. 

THEOREM 3.6 If /: X C R 2 R is differentiable at (a, b), then it is continu- 
ous at (a, b). 



2.3 | The Derivative 123 



EXAMPLE 7 Let the function /: R 2 

/(*. y) -- 



2 2 

x y 



R be denned by 

if(jc, 30^(0,0) 



x 4 + y 4 

0 if(x,y) = (0,0) 

The function / is not continuous at the origin, since lim^ ^j^^o.o) fi x i y) does 
not exist. (However, / is continuous everywhere else in R .) By Theorem 3.6, / 
therefore cannot be differentiate at the origin. Nonetheless, the partial derivatives 
of / do exist at the origin, and we have 



and 



fix, 0) = 



7(0, .v) = 



o 



x 4 + 0 



0+ v 4 



0 



0 



df 

dx 



df 
dy 



(0,0) = 0, 



(0,0) = 0, 



since the partial functions are constant. Thus, we see that if we want something 
like Theorem 3.6 to be true, the existence of partial derivatives alone is not 
enough. ♦ 



Differentiability in General 

It is not difficult now to see how to generalize Definition 3.4 to three (or more) 
variables: For a scalar-valued function of three variables to be differentiable at a 
point (a, b, c), we must have that (i) the three partial derivatives exist at (a, b, c) 
and (ii) the function h : R 3 — > R defined by 

h(x, y, z) = f(a, b, c) + f x {a, b, c)(x — a) 

+ f y (a, b, c)(y -b) + f t (a, b, c)(z - c) 

is a good linear approximation to / near (a, b, c). In other words, (ii) means that 

f(x,y,z)-h(x,y,z) 
hm = 0. 

(x,y,zy+(a,b,c) ||(x, V, z) ~ (a, b, C)\\ 

The passage from three variables to arbitrarily many is now straightforward. 



DEFINITION 3.7 Let X be open in R" and /: X R be a scalar-valued 
function; let a = (a\, a%, . . . , a„) e X. We say that / is differentiable at 
a if all the partial derivatives f Xj (a), i = 1, . . . , n, exist and if the function 
h:R n -> R defined by 

H*) = /(a) + Ai(a)(xi - fli) + - a 2 ) 

H h f Xn {&)(x n - a n ) (6) 

is a good linear approximation to / near a, meaning that 

hm = 0. 

*->■» II x — all 



1 24 Chapter 2 | Differentiation in Several Variables 



We can use vector and matrix notation to rewrite things a bit. Define the 
gradient of a scalar- valued function /: X C R" R to be the vector 

df_ df_ df_ 



V/(x) = 
Consequently, 



V/(a) = (A- 1 (a),/ X2 (a),..., A„(a)). 

Alternatively, we can use matrix notation and define the derivative of / at a, 
denoted Df(a), to be the row matrix whose entries are the components of V /(a); 
that is, 

»/(») = [/*(») M») ■■■ /,„(a)]. 

Then, by identifying the vector x — a with the n x 1 column matrix whose entries 
are the components of x — a, we have 



V/(a).(x-a) = D/(a)(x-a) = [/ Tl (a) /, 2 (a) ••• A„(a)] 



X\ — d\ 

x 2 - a 2 



= /xi(a)(*i - fli) + /x 2 (a)(x 2 - a 2 ) 

H 1- /v„(a)(x„ - a n ). 

Hence, vector notation allows us to rewrite equation (6) quite compactly as 
ft(x)=/(a) + V/(a).(x-a). 

Thus, to say that /z is a good linear approximation to / near a in equation (6) 
means that 

||x - a|| 

Compare equation (7) with equation (5). Differentiability of functions of one and 
several variables should really look very much the same to you. It is worth noting 
that the analogues of Theorems 3.5 and 3.6 hold in the case of n variables. 

The gradient of a function is an extremely important construction, and we 
consider it in greater detail in §2.6. 

You may be wondering what, if any, geometry is embedded in this general 
notion of differentiability. Recall that the graph of the function /:lcR"->R 
is the hypersurface in R" +1 given by the equation x n +\ = f(xi, x 2 , ... , x n ). 
(See equation (2) of §2.1.) If / is differentiable at a, then the hypersurface deter- 
mined by the graph has a tangent hyperplane at (a, /(a)) given by the equation 

x„ + i = h(x { ,x 2 ,...,x„) = /(a) + V/(a) • (x - a) 

= /(a) + £>/(a)(x - a). (8) 

Compare equation (8) with equation (3) for the tangent line to the curve y = F(x) 
at (a, F(a)). Although we cannot visualize the graph of a function of more than 
two variables, nonetheless, we can use vector notation to lend real meaning to 
tangency in n dimensions. 



EXAMPLE 8 Before we drown in a sea of abstraction and generalization, let's 
do some concrete computation. An example of an "« -dimensional paraboloid" in 



2.3 | The Derivative 1 25 



R n+1 is given by the equation 



X n +\ — X\ + *2 + ■ ■ ■ + X 2 , 



that is, by the graph of the function /(xi, . . . ,x„) = x 2 + x| H + x 2 . We have 

— = 2x ; - , z = 1, 2, . . . , n, 

dxi 

so that 



V/(*i 



x„) = (2xi, 2x 2 , 



2x„). 



Note that the partial derivatives of / are continuous everywhere. Hence, the 
n -dimensional version of Theorem 3.5 tells us that / is differentiate everywhere. 
In particular, / is differentiable at the point (1,2, ...,«), 



V/(l,2, ...,n) = (2,4, ...,2n), 



and 



Df(l,2,...,n)= [2 4 ■■■ In]. 
Thus, the paraboloid has a tangent hyperplane at the point 



(1,2, \ 2 + 2 2 + ---+n 2 ) 



whose equation is given by equation (8): 



x, 1+1 =(l 2 + 2 2 + 



= (l 2 + 2 2 + 



+ n 2 )+[2 4 ■■■ 2n] 



X\ — 1 
x 2 - 2 



+ n 2 ) + 2(xi - 1) + 4(x 2 -!) + ■■■+ 2n{x n - n) 



= (l 2 + 2 2 + 



+ n 2 ) + 2x] + 4x 2 + • 
(2 ■ 1 + 4 ■ 2 H \-2n-n) 



= 2xi + 4x 2 H h 2«x„ - (l 2 + 2 2 + 



2«x„ 



+ « 2 ) 



= 2i'x, 



n(n + l)(2n + 1) 



(The formula l 2 + 2 2 + ■ ■ ■ + n 2 = n(n + l)(2n + l)/6 is a welhlmown identity, 
encountered when you first learned about the definite integral. It's straightforward 
to prove using mathematical induction.) ♦ 

At last we're ready to take a look at differentiability in the most general setting 
of all. Let X be open in R" and let f: X —> R m be a vector- valued function of n 
variables. We define the matrix of partial derivatives of f, denoted Df, to be 



Chapter 2 | Differentiation in Several Variables 



the m x n matrix whose ijth entry is df/dxj, where f:X C R" 
component function of f. That is, 



R is the (th 



Df(x\,x 2 , ...,x n ) = 



dxi 
dx\ 



dxi 



dX2 

d_h 

dx 2 



Mm 
dX2 



dx n 

M. 

dx„ 



Mm 

dx n 



The ith row of Df is nothing more than Dfj — and the entries of Df are precisely 
the components of the gradient vector V f, (Indeed in the case where m = 1, 
V/ and Df mean exactly the same thing.) 

EXAMPLE 9 Suppose f: R 3 -> R 2 is given by f(x, y, z) = (x cos y + z, xj). 
Then we have 



Df(x,y,z) = 



cos y —x sin y 

y ^ 



We generalize equation (7) and Definition 3.7 in an obvious way to make the 
following definition: 



DEFINITION 3.8 (Grand definition of differentiability) Let X c 
R" be open, let f : X ->• R", and let a e X. We say that f is differentiable at 
a if Df(a) exists and if the function h: R" -> R" defined by 

h(x) = f(a) + Df(a)(x - a) 

is a good linear approximation to f near a. That is, we must have 



lim 



II fW - h(x) 



lim 



|f(x)-[f(a) + Df(a)(x-a) 



= 0. 



Some remarks are in order. First, the reason for having the vector length 
appearing in the numerator in the limit equation in Definition 3.8 is so that there 
is a quotient of real numbers of which we can take a limit. (Definition 3 . 7 concerns 
scalar-valued functions only, so there is automatically a quotient of real numbers.) 
Second the term Df(a)(x — a) in the definition of h should be interpreted as the 
product of the m x n matrix Df(a) and the nxl column matrix 



x 2 



Q2 



Because of the consistency of our definitions, the following results should 
not surprise you: 



2.3 | The Derivative 1 27 



THEOREM 3.9 If f: X C R" R m is differentiable at a, then it is continuous 
at a. 



THEOREM 3.10 If f:lcR"^ R"' is such that, for i = l,...,m and 
j = 1, . . . , n, all dfi/dxj exist and are continuous in a neighborhood of a in 
X, then f is differentiable at a. 



THEOREM 3.1 1 A function f: X C R" -> R m is differentiable ataeX (in the 
sense of Definition 3.8) if and only if each of its component functions fi'.X C 
R" — » R, i' = 1, . . . , m, is differentiable at a (in the sense of Definition 3.7). 

The proofs of Theorems 3.9, 3.10, and 3.11 are provided in the addendum 
to this section. Note that Theorems 3.10 and 3.1 1 frequently make it a straight- 
forward matter to check that a function is differentiable: Just look at the partial 
derivatives of the component functions and verify that they are continuous. Thus, 
in many — but not all — circumstances, we can avoid working directly with the 
limit in Definition 3.8. 



EXAMPLE 10 The function g: R 3 - {(0, 0, 0)} R 3 given by 

3 



g(*> y,z) 



x 2 + y 2 + z 2 



has 



Dg(x, y, z) = 



-6z 



—6x —6y 

(x 2 + y 2 + z 2 ) 2 (x 2 + y 2 + z 2 ) 2 (x 2 + y 2 + z 2 ) 2 
y x 0 

z 0 x 



Each of the entries of this matrix is continuous over R 3 — {(0, 0, 0)}. Hence, by 
Theorem 3. 10, g is differentiable over its entire domain. ♦ 



What Is a Derivative? 

Although we have defined quite carefully what it means for a function to be 
differentiable, the derivative itself has really taken a "backseat" in the preceding 
discussion. It is time to get some perspective on the concept of the derivative. 

In the case of a (differentiable) scalar-valued function of a single variable, 
/:XcR->R, the derivative f'ia) is simply a real number, the slope of the 
tangent line to the graph of / at the point (a, f(a)). From a more sophisticated 
(and slightly less geometric) point of view, the derivative f'{a) is the number such 
that the function 

h(x) = /(c) + f'(a)(x - a) 

is a good linear approximation to f(x) for x near a. (And, of course, y = h(x) is 
the equation of the tangent line.) 

If a function /:XcR"-^Rof« variables is differentiable, there must 
exist n partial derivatives 3 f/dx\ , . . . , df/dx„ . These partial derivatives form the 
components of the gradient vector V/ (or the entries of the 1 x n matrix Df). It 



Chapter 2 | Differentiation in Several Variables 



is the gradient that should properly be considered to be the derivative of /, but in 
the following sense: V/(a) is the vector such that the function h: R" —> R given 
by 

fc(x)=/(a) + V/(a).( X -a) 

is a good linear approximation to /(x) for x near a. Finally, the derivative of a 
differentiable vector- valued function f : X c R" — > R" ! may be taken to be the 
matrix Df of partial derivatives, but in the sense that the function h: R" — > R" 1 
given by 

h(x) = f(a) + Df(a)(x - a) 

is a good linear approximation to f(x) near a. You should view the derivative 
Df(a) not as a "static" matrix of numbers, but rather as a matrix that defines a 
linear mapping from R" to R'". (See Example 5 of §1.6.) This is embodied in 
the limit equation of Definition 3.8 and, though a subtle idea, is truly the heart of 
differential calculus of several variables. 

In fact, we could have approached our discussion of differentiability much 
more abstractly right from the beginning. We could have defined a function f:lc 
R" — > R°' to be differentiable at a point a 6 X to mean that there exists some 
linear mapping L: R" — > R'" such that 

Hm ||f(x)-[f(a) + L(x-a)]|| = Q 
x^a ||x — a|| 

Recall that any linear mapping L: R" — > R'" is really nothing more than multipli- 
cation by a suitable m x n matrix A (i.e., that L(y) = Ay). It is possible to show 
that if there is a linear mapping that satisfies the aforementioned limit equation, 
then the matrix A that defines it is both uniquely determined and is precisely the 
matrix of partial derivatives Df(a). (See Exercises 60-62 where these facts are 
proved.) However, to begin with such a definition, though equivalent to Definition 
3.8, strikes us as less well motivated than the approach we have taken. Hence, we 
have presented the notions of differentiability and the derivative from what we 
hope is a somewhat more concrete and geometric perspective. 



Addendum: Proofs of Theorems 3.9, 3.10, and 3.11 

Proof of Theorem 3.9 We begin by claiming the following: Let xeR" and 
B = (bjj) be an m x n matrix. If y = Bx, (so y 6 R m ), then 

llyll <TOI, (9) 

/ ,\l/2 

where K = ( >_\ . bf- 1 . We postpone the proof of (9) until we establish the 
main theorem. 

To show that f is continuous at a, we will show that ||f(x) — f(a)|| —> 0 as 
x -> a. We do so by using the fact that f is differentiable at a (Definition 3.8). 
We have 

||f(x) - f(a)|| = ||f(x) - f(a) - Df(a)(x - a) + Df(a)(x - a)|| 

< ||f(x) - f(a) - £>f(a)(x - a)|| + ||Df(a)(x - a)||, (10) 

using the triangle inequality. Note that the first term in the right side of inequality 
(10) is the numerator of the limit expression in Definition 3.8. Thus, since f is 



2.3 | The Derivative 1 29 



differentiable at a, we can make ||f(x) — f(a) — £>f(a)(x — a)|| as small as we wish 
by keeping ||x — a|| appropriately small. In particular, 

||f(x)-f(a)-Df(a)(x-a)|| < ||x-a|| 

if || x — a|| is sufficiently small. To the second term in the right side of inequality 
(10), we may apply (9), since Df(a) is an m x n matrix. Therefore, we see that if 
|| x — a|| is made sufficiently small, 

||f(x)-f(a)|| < ||x-a||+Z||x-a||=(l + Z)||x-a||. (11) 

The constant K does not depend on x. Thus, as x —> a, we have 

||f(x) - f(a)|| -» 0, 

as desired. 

To complete the proof, we establish inequality (9). Writing out the matrix 
multiplication, 



y = Bx = 



b\\%\ + ^12^2 H V b Xn x n 

bi\x\ + b 22 x 2 H h b 2n x„ 

b m \X\ + b m 2X2 + ■■■ + & 



bi • x 
b 2 -x 



where b, denotes the ith row of B, considered as a vector in R". Therefore, using 
the Cauchy-Schwarz inequality, 



||y|| = ((bi • X) 2 + (b 2 • X) 2 + • • • + (b m • X) 2 ) 
< (||b 1 || 2 ||x|| 2 + ||b2|r||x|r + 

,2\ 1/2 



+ l|b m || 2 ||x|| 2 ) 1/2 



|b 1 || 2 + ||b 2 || 2 + --- + ||b„ 



2 )' 



Now, 



Consequently, 



= bj l +bf 2 + ---+bl = J2bf J . 



llbi || 2 + ||b 2 1| 2 + • • • + ||b„, || 2 = lib, II 2 = E E b l = Rl - 

i=i i=i j=\ 

Thus, ||y|| < AT||x||, and we have completed the proof of Theorem 3.9. ■ 

Proof of Theorem 3.10 First, we prove Theorem 3. 10 for the case where / is a 
scalar-valued function of two variables. We begin by writing 

f(x\ , x 2 ) — f(flu a 2 ) = f(x\ , xi) - f{a\ , x 2 ) + f(a\ , x 2 ) - f(a\, a 2 ). 

By the mean value theorem, 2 there exists a number c\ between a\ and xi such 

that 



f{x\,x 2 ) - f(ai,x 2 ) = f Xi (c\, x 2 ){x x - ai) 



2 Recall that the mean value theorem says that if F is continuous on the closed interval [a , b] and differen- 
tiable on the open interval (a, b), then there is a number c in (a , b) such that F(b) — F(a) = F'(c)(b — a). 



Chapter 2 | Differentiation in Several Variables 



and a number c 2 between a 2 and x 2 such that 

f(ai,x 2 ) - f(a u a 2 ) = f X2 (a u c 2 )(x 2 - a 2 ). 

(This works because in each case we hold all the variables in / constant except 
one, so that the mean value theorem applies.) Hence, 

\f(x u x 2 ) - f(a\, a 2 ) - f xi {a\, a 2 ){x\ - a{) - f X2 (a u a 2 )(x 2 - a 2 )\ 
= \f Xl (c\,x 2 )(x\ - ai)+ f X2 (ai, c 2 )(x 2 - a 2 ) - f X] (a u a 2 )(x\ - a x ) 
-f X2 (a u a 2 )(x 2 - a 2 )\ 

< x 2 )(.x;i - ai) - f Xl (a x ,a 2 )(xi - fli)| 

+ |/x 2 (ai, c 2 )(x 2 - a 2 ) - f Xl (ai, a 2 )(x 2 - a 2 )\ , 
by the triangle inequality. Hence, 

\f(xi,x 2 ) - f(a u a 2 ) - f xi (ai,a 2 )(xi - a\) - f X2 (a u a 2 )(x 2 - a 2 )\ 

< |/*i(Cl,*2) ~ fxM\' a 2)\ \ x \ ~ a \\ 

+ \f X2 (a u c 2 ) - f X2 {a x , a 2 )\ \x 2 - a 2 \ 

< {\f Xl (ci,x 2 ) - f Xl (ai,a 2 )\ + \f X2 (ai, c 2 ) - f X2 (a u a 2 )\) ||x - a||, 
since, for i = 1,2, 

\ Xi -en\< ||x - a|| = ((xi - a x ) 2 + (x 2 - a 2 ) 2 ) 1 ' 2 . 

Thus, 

|/(xi, x 2 ) - f(a u a 2 ) - f xi (a u a 2 )(xi - a { ) - f X2 {a x ,a 2 ){x 2 - a 2 )\ 

l|x-a|| 

< \f xi (c\,x 2 )- f xl (ai,a 2 )\ + \f X2 (a u c 2 )- f X2 (ai,a 2 )\. (12) 

As x — > a, we must have that c, —> a\ , for i = I, 2, since c, is between a, and x, . 
Consequently, by the continuity of the partial derivatives, both terms of the right 
side of (12) approach zero. Therefore, 

,. \f(xi,x 2 )- f(a x ,a 2 )- f Xl (ai,a 2 )(xi - a{) - f X2 (a { , a 2 )(x 2 - a 2 )\ 

hm 1 = 0 

||x-a|| 

as desired. 

Exactly the same kind of argument may be used in the case that / is a scalar- 
valued function of n variables — the details are only slightly more involved so 
we omit them. Granting this, we consider the case of a vector-valued function 
f: R" — > R m . According to Definition 3.8, we must show that 

||m l|f W -f(a)-Z>f W (x- a) | =0 
||x - a|| 

The component functions of the expression appearing in the numerator may be 
written as 

G, = Mx) - fi(z) - D/,(a)(x - a), (14) 

where fi, i = 1, . . . , m, denotes the ith component function of f. (Note that, by 
the cases of Theorem 3.10 already established, each scalar- valued function /, is 



2.3 I Exercises 



differentiable.) Now, we consider 

||f(x) - f(a) - Df(a)(x - a)|| \\(G U G 2 , . . . , G„ 



|x-a|| ||x -a|| 

_ (G] + Gl + --- + Gjy /2 
Hi -a|| 

< |Gil + |G 2 | + --- + |G m | 
l|x-a|| 

Igil ] \G 2 \ | | \G„ 



lix — a|| ||x — a|| ||x-a|| 

Asx^ a, eachterm |G, |/||x — a|| -> 0, by definition of G, in equation (14) and 
the differentiability of the component functions f\ of f . Hence, equation (13) holds 

and f is differentiable at a. (To see that (G 2 H h G 2 ) 1/2 < \G\\-\ hi G m \ , 

note that 

(IGH H h |G m |) 2 = |Gi| 2 H h |G,„| 2 

+ 2|Gi| |G 2 | +2|Gi| |G 3 | + ■ ■ ■ + 2|G M _i| |G„,| 
> |Gi| 2 + --- + |GJ 2 . 
Then, taking square roots provides the inequality.) ■ 

Proof of Theorem 3.11 In the final paragraph of the proof of Theorem 3. 10, we 
showed that 

||f(x)-f(a)-Pf(a)(x-a)|| < |G t | | |G 2 | | | |G,„| 



||x-a|| llx — a|| ||x-a|| l|x-a|| 

where G, = fi(x) — /,(a) — Z)/}(a)(x — a) as in equation (14). From this, it fol- 
lows immediately that differentiability of the component functions f\ , . . . , f m at 
a implies differentiability of fat a. Conversely, for i = 1, . . . , m, 

||f(x)-f(a)-Df(a)(x-a)|| = \\(G U G 2 , . . . , G m )|| > |G,| 



||x-a|| ||x-a|| ||x-a|| 

Hence, differentiability of f at a forces differentiability of each component 
function. ■ 



2.3 Exercises 



In Exercises 1—9, calculate df/dx and df/dy. 
1- f(x,y) = xy 2 +x 2 y 

2. f(x, y) = e^ 1 

3. f(x, y) = sinxy + cosjcy 



6. f(x, y) = \n(x 2 + y 2 ) 

7. f(x, y) = cosx 3 y 

' x ' 



8. f(x, y) = In 



4. f(x,y) . 



5. f{x,y) . 



^ - y 2 
\+x 2 + 3/ 

x 2 -y 2 
x 2 + y 2 



9. f(x, y) = xe y + y sin(x 2 + y) 

In Exercises 10—17, evaluate the partial derivatives dF/dx, 
dF/dy, and dF/dz for the given functions F. 

10. F(x,y,z) = x + 3y-2z 



Chapter 2 | Differentiation in Several Variables 



x — v 

11. F(x,y,z)= — ~ 

y + z 

12. F(x, y, z) = xyz 

13. F(x,y,z) = y / x 2 + y 2 + z 2 

14. F(x, y, z) = e ax cosfcy + e az sinbx 

-ic T7i ^ x + y + z 

15. F(x, y, z) = — — 

(1 + x 2 + y 2 + Z 2 ) 3/2 

16. F(x, y, z) = sinx 2 y 3 z 4 



17. F(x,y,z) 



x 3 + yz 



X 2 + Z 2 + 1 

FzW //ze gradient V/(a), where f and a are given in Exer- 
cises 18-25. 

18. /(x, y) = x 2 y + e y ' x , a = (1,0) 
x — y 



19. f(x,y) . 



-, a = (2,-1) 



.v 2 + y 2 + V 

20. /(x, y, z) = sinxyz, a = (jt, 0, jr/2) 

21. /(x, j, z) = xy + y cosz — * sinyz, 
a = (2,-l,jr) 

22. f(x, y) = c»> + In (x - y), a = (2, 1) 

23. /(x,y, z )=^±2, a = (3, -1,0) 

24. /(x,y,z) = coszln(x + y 2 ), a = (e, 0, jt/4) 

2 _ 2 

25. /(x,y,z) = / Z , ■ = (-1,2,1) 

y z + z z + 1 

In Exercises 26-33, find the matrix Df( a) of partial derivatives, 
where f and a are as indicated. 

26. f(x,y)=-, a = (3, 2) 

y 

27. f(x,y,z) = x 2 +xln(yz), a = (-3, e,e) 

28. f(x, y, z) = (2x - 3y + 5z, x 2 + y, In (yz)), 
a = (3, -1,-2) 

29. f(x, y, z) = (xyz, ^x 2 + y 2 + z 2 ^), 

a = (1,0, -2) 

30. f(f) = (r, cos2f, sin 5/), a = 0 

31. f(x, y, z, id) = (3x - 7y + z, 5x + 2z - Sw, 
y-\7z + 3w), a = (1,2, 3, 4) 

32. f(x, y) = (x 2 y, x + y 2 , cos jrxy), a = (2, -1) 

33. f(s, r) = (j 2 ,jf,r 2 ), a = (-1,1) 

Explain why each of the functions given in Exercises 34—36 is 
differentiable at every point in its domain. 

34. /(x, y) = xy — 7x 8 y 2 + cosx 
x + v + z 



35. /(x,y,z): 



36. f(x, y) : 



xy 



x l + y 4 y 



37. (a) Explain why the graph of z = x 3 — 7xy + e y has 

a tangent plane at (— 1 , 0, 0). 
(b) Give an equation for this tangent plane. 

38. Find an equation for the plane tangent to the graph of 
z = 4cosxy at the point (tt/3, 1, 2). 

39. Find an equation for the plane tangent to the graph of 
z = e x+y cosxy at the point (0, 1, e). 

40. Find equations for the planes tangent to z = 
x 2 — 6x + y 3 that are parallel to the plane 



4x - 12y + j 



7. 



41. Use formula (8) to find an equation for the hy- 
perplane tangent to the 4-dimensional paraboloid 
x 5 = 10 — (x\ + 3x| + 2x| + x|) at the point 
(2,-1, 1,3,-8). 

42. Suppose that you have the following information con- 
cerning a differentiable function /: 

/(2, 3) =12, /(1.98,3) = 12.1, f(2, 3.01) = 12.2. 

( a) Give an approximate equation for the plane tangent 
to the graph of / at (2, 3, 12). 

(b) Use the result of part (a) to estimate /(l .98, 2.98). 

In Exercises 43—45, (a) use the linear function h(\) in Def- 
inition 3.8 to approximate the indicated value of the given 
function f. (b) How accurate is the approximation determined 
in part (a)? 

43. f(x, y) = e x+y , /(0.1, -0.1) 

44. f(x, y) = 3 + cos itxy, /(0.98, 0.51) 

45. f(x, y, z) = x 2 + xyz + y 3 z, /(1.01, 1.95, 2.2) 

46. Calculate the partial derivatives of 

xi + Xj H 1- x„ 



/(Xl,X 2 , 



47. Let 



fix, y) 



xy 



, x„) 



■x 2 y + 3x 3 



*? + xl + ■ 



■+x, 



x 2 + y 2 
0 



if(x,y)^(0, 0) 
if(x,y) = (0,0) 



x 2 + y 2 + z 2 



(a) Calculate df/dx and df/dy for (x, y)/(0, 0). 
(You may wish to use a computer algebra system 
for this part.) 

(b) Find f x (0, 0) and /,(0, 0). 

As mentioned in the text, if a function F{x) of a single variable 
is differentiable at a, then, as we zoom in on the point {a, F(a)), 
the graph of y = F(x) will "straighten out" and look like its 
tangent line at (a, F(a)). For the differentiable functions given 



2.3 I Exercises 133 



in Exercises 48-51, (a) calculate the tangent line at the indi- 
cated point, and (b) use a computer to graph the function and 
the tangent line on the same set of axes. Zoom in on the point 
of tangency to illustrate how the graph of y = F(x) looks like 
its tangent line near (a, F(a)). 



^ 48. F(x) 



2x + 3, a = 1 



smx, a 



^ 49. F{x) = x + - 

>- 3 - 3x 2 + x 



x —ox -(- X 

O 50. F(x) = T — , a = 0 

x L + 1 

^ 51. F(x) = ln(x 2 + 1), a = -1 

52. (a) Use a computer to graph the function F(x) = 
(x-2) 2 '\ 

(b) By zooming in near x = 2, offer a geometric dis- 
cussion concerning the differentiability of F at 
x =2. 

As discussed in the text, a function fix, y) may have partial 
derivatives f x (a, b) and f y (a, b) yet fail to be differentiable at 
(a, b). Geometrically, if a function fix, y) is differentiable at 
(a, b), then, aswezoom in on thepoint(a, b, f(a, b)), thegraph 
of z = fix, y) will "flatten out" and look like the plane given 
by equation (4) in this section. For the functions fix, y) given 
in Exercises 53-57, (a) calculate f x (a, b) and f y (a, b) at the 
indicated point (a , b) and write the equation for the plane given 
by formula (4) of this section, (b) use a computer to graph the 
equation z = f(x, y) together with the plane calculated in part 
(a). Zoom in near the point (a, b, f(a, b)) and discuss whether 
or not f(x, y) is differentiable at (a, b). (c) Give an analytic 
(i.e., nongraphical) argument for your answer in part (b). 

ft 53. f(x, y) = x 3 -xy + y 2 , (a, b) = (2, 1) 
^> 54. f{x, y) = Hx - \)yf'\ (a, b) = (1, 0) 



^55. fix,y) . 
^56. fix,y) . 



xy 



( fl ,fc) = (0,0) 



x 2 + y 2 + 1 ' 

fit 3tt \ 

sinxcosy, (a, &)=(—,— ) 
V 6 4 / 



^ 57. fix,y) = x 2 s'my + y 2 cosx, (a,A)=^— ,— ) 

58. Let gix, y) = ^fxy. 

(a) Is g continuous at (0, 0)? 

(b) Calculate dg/dx and dg/dy whenxy ^ 0. 

(c) Show that g x (0, 0) and g, (0, 0) exist by supplying 
values for them. 

(d) Are dg/dx and dg/dy continuous at (0, 0)? 

(e) Does the graph of z = gix, y) have a tangent plane 
at (0, 0)? You might consider creating a graph of 
this surface. 

(f) Is g differentiable at (0, 0)? 



59. Suppose f: R" —> W is a linear mapping; that is, 

f(x) = Ax, where x = (xi, X2, . . . , x n ) e R" 

and A is an m x n matrix. Calculate Df(x) and relate 
your result to the derivative of the one-variable linear 
function fix) = ax. 

In Exercises 60-62 you will establish that the matrix -Df(a) of 
partial derivatives of the component functions ofiis uniquely 
determined by the limit equation in Definition 3.8. 

60. Let X be an open set in R", let aeX, and let F:ic 
R" -± R". Show that 



lim ||F(x)|| = 0 



lim F(x) = 0. 



61. Let X be an open set in R", let ael, and let f:XC 
R" — >• R" . Suppose that A and B are m x n matrices 
such that 



lim 

x— >a 



lim 



||f(x)-[f(a) + A(x-a)]| 

ll x — a|| 
f(x) - [f(a) + 5(x - a)] || 



x^a || x — a|| 

(a) Use Exercise 60 to show that 

hmCS-AKx-a, 
x->-a Hx — all 



0. 



(b) Write x — a as fh, where h is a nonzero vector in 
R" . First argue that 



lim 

x-»a 



(S - A)(x - a) 



0 implies 



lim (B-A)W 
t-o \\th\\ 

and then use this result to conclude that A = B. 
(Hint: Break into cases where t > 0 and where 
t < 0.) 

62. Let X be an open set in R", let aeX, and let f:XC 
R" — >• R" . Suppose that A is an m x n matrix such 
that 



lim 

x->a 



||f(x)-[f(a) + A(x-a)] 



0. 



In this problem you will establish that A = Df(a). 
(a) Define F:ICR"-> R" by 



F(x) = 



f(x) - f(a) - A(x - a) 



Identify the ;th component function us- 
ing component functions of f and parts of the 
matrix A. 

(b) Note that under the assumptions of this problem 
and Exercise 60, we have that lim x ^ a F(x) = 0. 



1 34 Chapter 2 | Differentiation in Several Variables 



First argue that, for i=l,...,m, we have 
lim x ^ a Fi(x) = 0. Next, argue that 

lim F,(x) = 0 implies lim F,-(a + he:) = 0, 

where denotes the standard basis vector 
(0, 1, ...,0)forR". 



(c) Use parts (a) and (b) to show that a t j 



M. 



(a), 



where a, ; denotes the ijth entry of A. (Hint: Break 
into cases where h > 0 and where h < 0.) 



2.4 Properties; Higher-order Partial 
Derivatives 

Properties of the Derivative 

From our work in the previous section, we know that the derivative of a function 
f : X c R" R'" can be identified with its matrix of partial derivatives. We next 
note several properties that the derivative must satisfy. The proofs of these results 
involve Definition 3.8 of the derivative, properties of ordinary differentiation, and 
matrix algebra. 



PROPOSITION 4.1 (Linearity of differentiation) Let f,g:Jfc R" -» R ffl 
be two functions that are both differentiable at a point aeX, and let c e R be 
any scalar. Then 

1. The function h = f + g is also differentiable at a, and we have 

Z)h(a) = D(f + g)(a) = Df(a) + £»g(a). 

2. The function k = cf is differentiable at a and 

£>k(a) = £>(cf)(a) = c£>f(a). 



EXAMPLE 1 Let f and g be defined by f(x, y) = (x + y, xy siny, y/x) and 
g(x, y) = (x 2 + y 2 , ye xy , 2x 3 — 7y 5 ). We have 



Df(x, y) = 



1 

y siny 

L -3V* 2 



and 



Dg(x,y) = 



2x 

y 2 e xy 
6x 2 



1 

x siny + xy cosy 
l/x 



2y 

e x> ' + xye^^ 
-35y 4 



Thus, by Theorem 3.10, f is differentiable on R 2 — {y-axis} and g is differentiable 
on all of R 2 . If we let h = f + g, then part 1 of Proposition 4. 1 tells us that h must 
be differentiable on all of its domain, and 

Dh(x,y) = Df(x,y) + Dg(x,y) 

2x + \ 2y + 1 

y sin y + y 2 e xy x sin y + xy cos y + e xy + xye xy 
6x 2 -y/x 2 l/x- 35y 4 



2.4 | Properties; Higher-order Partial Derivatives 



Note also that the function k = 3g must be differentiable everywhere by part 2 
of Proposition 4.1. We can readily check that Dk(x, y) = 3Dg(x, y): We have 

k(x, y) = (3x 2 + 3y 2 , 3ye xy , 6x 3 - 21y 5 ). 

Hence, 



£>k(x, y) = 



6.v 


6y 




3e xy + 3xye 


18x 2 


-105y 4 


2x 


2y 


y 2 e xy 


e xy + xye xy 


6x 2 


-35y 4 



= 3 



= 3£>g(x, y). 4 

Due to the nature of matrix multiplication, general versions of the product 
and quotient rules do not exist in any particularly simple form. However, for 
scalar-valued functions, it is possible to prove the following: 

PROPOSITION 4.2 Let /,g:lcR"^Rbe differentiable ataeX Then 

1. The product function fg is also differentiable at a, and 

D(fgM = s(a)D/(a) + /(a)Z)g(a). 

2. If g(a) / 0, then the quotient function f/g is differentiable at a, and 

g(a)D/(a) - /(a)Z)g(a) 



D(//g)(a) = 



g(a) 2 



(/#)(*> y, z) = (xyz + 2yz 2 - xz 2 )e xy , 



EXAMPLE 2 If f(x, y, z) = z<? x - v and g(x, y, z) = xy + 2yz - xz, then 
so that 

D(fg)(x,y,z) 



(yz - z 2 )e xy + (xyz + 2yz 2 - xz 2 )ye xy 
(xz + 2z 2 )e xy + (xyz + 2yz 2 - xz 2 )xe xy 
(xy + 4yz — 2xz)e x> ' 



-i t 



Also, we have 
and 



Df(x,y,z)= [yze xy x Z e xy e xy ] 
Dg(x,y,z)=[y - z x + 2z 2y-x], 



so that 



g(x, y, z)Df(x, y, z) + f(x, y, z)Dg(x, y, z) 

(xy 2 z + 2y 2 z 2 — xyz 2 )e xy 
(x 2 yz + 2xyz 2 - x 2 z 2 )e xy 
(xy + 2yz - xz)e xy 



(yz - z 2 )e xy 
(xz + 2z 2 )e xv 
(2yz - xz)e xy 



-1 T 



Chapter 2 | Differentiation in Several Variables 



= e 



xy 2 z 
x 2 yz - 



- 2y 2 z 2 - xyz 2 + yz- z 2 
2xyz 2 - x 2 z 2 + xz + 2z 2 
xy + 4yz — 2xz 



which checks with part 1 of Proposition 4.2. (Note: The matrix transpose is used 
simply to conserve space on the page.) ♦ 



The product rule in part 1 of Proposition 4.2 is not the most general 
result possible. Indeed, if /:ICR"->R is a scalar- valued function and 
g:lcR"^ R'" is a vector- valued function, then if / and g are both differ- 
entiate at a e X, so is fg, and the following formula holds (where we view g(a) 
as an m x 1 matrix): 

£>(/g)(a) = g(a)D/(a) + /(a)Dg(a). 



Partial Derivatives of Higher Order 

Thus far in our study of differentiation, we have been concerned only with partial 
derivatives of first order. Nonetheless, it is easy to imagine computing second- 
and third-order partials by iterating the process of differentiating with respect to 
one variable, while all others are held constant. 

EXAMPLE 3 Let f(x, y, z) = x 2 y + y 2 z. Then the first-order partial deriva- 
tives are 

9/ „ 9 / 2 „ 9/ 2 
— = 2xy, — = x + 2yz, and — = y , 

ax ay az 

The second-order partial derivative with respect to x, denoted by d 2 f/dx 2 or 
f xx (x, y, z), is 

d 2 f 9 (df\ 3 .„ , „ 

— — = — — = — ( 2x y) = 2 y- 

dx 2 dx \dx ) dx 
Similarly, the second-order partials with respect to y and z are, respectively, 

9 2 / 9 /3A d 



dy 2 dy \dy J dy^ 



and 

d 2 f 3 (df 



dz 2 dz \ dz J dz ^ 

There are more second-order partials, however. The mixed partial derivative 
with respect to first x and then y, denoted d 2 f /dydx or f xy (x, y, z), is 

9 2 / 9 /3/\ 3 , 

= — (2xy) = 2x. 



dydx dy \dx ) dy 

There are five more mixed partials for this particular function: d 2 f/dxdy, 
d 2 f/dzdx, d 2 f /dxdz, d 2 f /dzdy, and d 2 f /dydz. Compute each of them to get 
a feeling for the process. ♦ 

In general, if /: X C R" R is a (scalar- valued) function of n variables, the 
&th-order partial derivative with respect to the variables (in that 



2.4 | Properties; Higher-order Partial Derivatives 



order), where ii , 1*2, -..,£* are integers in the set {1, 2, .... n} (possibly repeated), 
is the iterated derivative 

dx h - ■ ■ dx h dx il dx k dx k dx il 

Equivalent (and frequently more manageable) notation for this /rth-order partial 
is 

Note that the order in which we write the variables with respect to which we 
differentiate is different in the two notations: In the subscript notation, we write 
the differentiation variables from left to right in the order we differentiate, while 
in the 3 -notation, we write those variables in the opposite order (i.e., from right 
to left). 

EXAMPLE 4 Let f(x, y, z, w) = xyz + xy 2 w — cos(x + zw). We then have 

d 2 f 9 9 , 
f yw (x,y,z,w)= — — = — — (xyz + xy w - cos(x + zw)) 
away aw ay 

9 

= — (xz + 2xyw) = 2xy, 
dw 



and 



d 2 f 9 9 2 
f wy (x, y, z, w) = —— = — —(xyz + xy w - cos(x + zw)) 
ayaw ay aw 



9 

9^ 



(xy 2 + z sin(x + zw)) = 2xy. 



♦ 



Although it is generally ill-advised to formulate conjectures based on a single 
piece of evidence, Example 4 suggests that there might be an outrageously simple 
relationship among the mixed second partials. Indeed, such is the case, as the next 
result, due to the 1 8th-century French mathematician Alexis Clairaut, indicates. 



THEOREM 4.3 Suppose that X is open in R" and /:XcR"^R has con- 
tinuous first- and second-order partial derivatives. Then the order in which we 
evaluate the mixed second-order partials is immaterial; that is, if i\ and 12 are any 
two integers between 1 and n, then 

J 2 J_ = J 2 J_ 
dxj 1 dxj 2 dx i2 dxj 1 



A proof of Theorem 4.3 is provided in the addendum to this section. We 
also suggest a second proof (using integrals!) in Exercise 4 of the Miscellaneous 
Exercises for Chapter 5. 

It is natural to speculate about the possibility of an analogue to Theorem 4.3 
for &th-order mixed partials. Before we state what should be an easily anticipated 
result, we need some terminology. 



Chapter 2 | Differentiation in Several Variables 



DEFINITION 4.4 Assume X is open in R". A scalar-valued function 
/:XcR"^R whose partial derivatives up to (and including) order at 
least k exist and are continuous on X is said to be of class C k . If / has 
continuous partial derivatives of all orders on X, then / is said to be of class 
C°°, or smooth. A vector-valued function f: X c R" -> R'" is of class C k 
(respectively, of class C°°) if and only if each of its component functions is 
of class C k (respectively, C°°). 



THEOREM 4.5 Let /: X c R" -> R be a scalar-valued function of class C k . 
Then the order in which we calculate any £th-order partial derivative does not 
matter: If . . . , /<-) are any k integers (not necessarily distinct) between 1 and 
n, and if (ji, ... , jk) is any permutation (rearrangement) of these integers, then 

d k f = d k f 
dxj { ■ ■ ■ dxj t dxjj ■ ■ ■ dxj t 



EXAMPLE 5 If f(x, y, z, w) = x 2 we yz — ze xw + xyzw, then you can check 
that 



9 5 f d 5 f 
2e yz (yz + \) 



dxdwdzdydx dzdydwd 2 x' 
verifying Theorem 4.5 in this case. 



Addendum: Two Technical Proofs 

Proof of Part 1 of Proposition 4.1 

Step 1. We show that the matrix of partial derivatives of h is the sum of 
those of f and g. If we write h(x) as (hi(x), /i2(x), . . . , h m (x)) (i.e., in terms of 
its component functions), then the ijth entry of £>h(a) is dhj/dxj evaluated at a. 
But hj(x) = fi(x) + gi(x) by definition of h. Hence, 

dh; 9 dfj dg; 

— ^ = —(/;■« + oo) = -f- + 

dXj dXj axj oxj 

by properties of ordinary differentiation (since all variables except xj are held 
constant). Thus, 

^— (a)= — (a)+ — (a), 

axj axj axj 



and therefore, 



Z)h(a) = £>f(a) + Dg(a). 



Step 2. Now that we know the desired matrix of partials exists, we must 
show that h really is differentiable; that is, we must establish that 

Hm ||h(x)-[h(a) + Ph(a)(x-a)]|| = Q 



2.4 | Properties; Higher-order Partial Derivatives 



(a, b + Ay) 


(a + Ax, b + Ay) 


• 


• 

+ 


+ 






• 


(a'b) 


(a + Ax, b) 



Figure 2.55 To construct the 
difference function D used in the 
proof of Theorem 4.3, evaluate / 
at the four points shown with the 
signs as indicated. 



As preliminary background, we note that 

||h(x)-[h(a) + /)h(a)(x-a)]|| 
l|x-a|| 

||f(x) + g(x) - [f(a) + g(a) + £>f(a)(x - a) + Dg(a)(x - a)]|| 

II* -a|| 

||(f(x) - [f(a) + Df(a)(x - a)]) + (g(x) - [g(a) + Dg(a)(x - a)])|| 

II* -a|| 

< ||f(x)-[f(a) + Df(a)(x-a)]|| | ||g(x) - [g(a) + Dg(a)(x - a)]|| 



x-a 



x - a 



by the triangle inequality, formula (2) of §1.6. To show that the desired limit 
equation for h follows from the definition of the limit, we must show that given 
any e > 0, we can find a number S > 0 such that 



., n , , . t , ||h(x)-[h(a) + £>h(a)(x-a)]|| 

if 0 < || x - a|| < 8, then < e. 



x - a 



(1) 



Since f is given to be differentiable at a, this means that given any e\ > 0, we can 
find S\ > 0 such that 



•fn ,i ,i * a llf(x)-[f(a) + flf(a)(x-a)]| 
if 0 < || x — a|| < Si, then < ei. 



x-a 



(2) 



Similarly, differentiability of g means that given any 62 > 0, we can find a 82 > 0 
such that 



Tn „ „ . t , ||g(x)-[g(a) + J Dg(a)(x-a)]| 
if 0 < || x — a|| < h, then < €2. 



x - a 



(3) 



Now we're ready to establish statement (1). Suppose e > 0 is given. Let 81 
and 8 2 be such that (2) and (3) hold with ei = €2 = e/2. Take 8 to be the smaller 
of 81 and 82- Hence, if 0 < ||x — a|| < <5, then both statements (2) and (3) hold 
(with €1 = €2 = e/2) and, moreover, 

||h(x)-[h(a) + Ph(a)(x-a)]|| ||f(x) - [f(a) + Df(a)(x - a)]|| 



x-a 



Hi -a|| 

|g(x)-[g(a) + Z)g(a)(x-a)]| 



+ 



x-a 



<e l +e 2 
e e 
= 2 + 2= € - 

That is, statement (1) holds, as desired. ■ 

Proof of Theorem 4.3 For simplicity of notation only, we'll assume that / is a 
function of just two variables (x and y). Let the point (a , b) e R 2 be in the interior 
of some rectangle on which f x , f y , f xx , f yy , f xy , and f yx are all continuous. 
Consider the following "difference function." (See Figure 2.55.) 

D(Ax, Ay) = f(a + Ax, b + Ay) - f(a + Ax, b) 
-f(a,b+ Ay) + f(a,b). 



1 40 Chapter 2 | Differentiation in Several Variables 



(a, b + Ay) (a + Ax, b + Ay) 





R 


{'d 





X 



(a,b) 



(a + Ax, b) 



Figure 2.56 Applying the mean 
value theorem twice. 



Our proof depends upon viewing this function in two ways. We first regard D as 
a difference of vertical differences in /: 

D(Ax, Ay) = [f(a + Ax,b + Ay) - f{a + Ax, b)] 

- [f(a, b + Ay) - f (a, b)] 

= F(a + Ax) - F(a). 

Here we define the one-variable function F(x) to be f(x, b + Ay) — f(x, b). As 
we will see, the mixed second partial of / can be found from two applications of 
the mean value theorem of one-variable calculus. Since / has continuous partials, 
it is differentiable. (See Theorem 3. 10.) Hence, F is continuous and differentiable, 
and, thus, the mean value theorem implies that there is some number c between 
a and a + Ax such that 



D(Ax, Ay) = F(a + Ax) - F(a) = F'(c)Ax. 



(4) 



Now F'{c) = f x {c, b + Ay) — f x (c, b). We again apply the mean value theorem, 
this time to the function f x (c, y). (Here, we think of c as constant and y as the 
variable.) By hypothesis f x is differentiable since its partial derivatives, f xx and 
f xy , are assumed to be continuous. Consequently, the mean value theorem applies 
to give us a number d between b and b + Ay such that 



F'(c) = f x (c, b + A.y) - f x (c, b) = f xy (c, d)Ay. 
Using equation (5) in equation (4), we have 

D(Ax, Ay) = F'(c)Ax = f xy (c, d)AyAx. 



(5) 



The point (c, d) lies somewhere in the interior of the rectangle R with vertices 
(a, b), {a + Ax, b), (a, b + Ay), (a + Ax, b + Ay), as shown in Figure 2.56. 
Thus, as (Ax, Ay) -> (0, 0), we have (c, d) — > (a, b). Hence, it follows that 



f xy (c, d) -> f xy (a, b) as (Ax, Ay) 
since f xy is assumed to be continuous. Therefore, 



(0, 0), 



D(Ax, Ay) 

f x Ja,b)= lim f x Jc,d)= lim . 

(Ajc,A)0-»-(o,O) " (Ax,Ajo^(0,0) AyAx 

On the other hand, we could just as well have written D as a difference of 
horizontal differences in /: 

£>(Ax, Ay) = [/(a + Ax, b + Ay) - f(a, b + Ay)] 

- [f(a + Ax, b) - f(a, b)] 

= G(b + Ay) - G(b). 

Here G(y) = f(a + Ax, y) — f(a, y). As before, we can apply the mean value 
theorem twice to find that there must be another point (c, d) in R such that 



D(Ax, Ay) = G'(d)Ay = f yx (c, d)AxAy. 



Therefore, 

f yx (a, b) = 



lim fvx(c, d) 

(Ax,Av)^(0,0) " 



lim 

(Ajt,Ay)->-(0,0) 



D(Ax, Ay) 
Ax Ay 



Because this is the same limit as that for f xy (a, b) just given, we have established 
the desired result. ■ 



2.4 I Exercises 141 



2.4 Exercises 



In Exercises 1—4, verify the sum rule for derivative matrices 
(i.e., part 1 of Proposition 4.1) for each of the given pairs of 
functions: 

1. f(x, y) — xy + cosx, g(x, y) = sin(xy) + y 3 

2. f{x, y) = (e x+y ,xe y ), g(x, y) = Qn(xy), ye x ) 

3. f(x, y, z) = (x sin y + z, ye z — 3x 2 ), g(x, y, z) = 
(x 3 cosx, xyz) 

4. f(x, y, z) = (xyz 2 , xe~ y , y sinxz), g(x, y, z) = 
(x — y, x 2 + y 2 + z 2 , ln(xz + 2)) 

Verify the product and quotient rules (Proposition 4.2) for the 
pairs of functions given in Exercises 5—8. 

x 

5- f(x, y) = x 2 y + y 3 , g(x, y) = - 

y 

6. f(x,y) = e xy , g(x, y) = x sin2y 

7- fix, y) = 3xy + y 5 , g(x, y) = x 3 - 2xy 2 

8. f(x,y,z) = xcos(yz), 

g(x, y, z) = x 2 + x 9 y 2 + y 2 z 3 + 2 

For the functions given in Exercises 9-1 7 determine all second- 
order partial derivatives (including mixed partials). 



9. 


fix, y) = 


x 3 y 7 + 3xy 2 — Ixy 


10. 


fix, y) = 


cos (xy) 


11. 


fix, y) = 


e y/x _ ye -x 


12. 


fix, y) = 


siny 7 * 2 + y 2 


13. 


fix, y) = 


1 


sin 2 x + 2e y 


14. 


fix,y) = 




15. 


fix, y) = 


y sin x — x cos y 


16. 


fix, y) = 




17. 


fix,y) = 


x 2 e y + e 2z 


18. 


fix, y, z) 


x — y 
y + z 


19. 


f(x,y,z) 


= x 2 yz + xy 2 z + xyz 2 


20. 


fix, y, z) 


= e xyz 


21. 


fix, y, z) 


= e ax sin y + e hx cos z 


22. 


Consider 


the function F(x,y,z) 



y 3 Z 5 — Ixyz. 



(a) Find F xx , F yy , and F- z . 

(b) Calculate the mixed second-order partials F xy , 
F yx , F xz , F^ x , F yz , and F zy , and verify Theorem 
4.3. 



(c) Is F xyx = F xxy l Could you have known this with- 
out resorting to calculation? 

(d) Is F xyz = F yzx 7 

23. Let f(x, y) = ye 3x . Give general formulas for 

d"f/dx n and d n f/dy n , where n > 2. 

24. Let f(x,y,z) = xe 2y + ye iz +ze~ x . Give general 
formulas for d"f/dx", d n f/dy n , and d"f/dz", where 
n > 1. 

25. Let f(x, y, z) = In ^— ^. Give general formulas for 

d"f/dx", d"f/dy n , and d n f/dz", where n > 1. What 
can you say about the mixed partial derivatives? 

26. Let f(x, y, z) = x 1 y 2 z 3 - 2x 4 yz. 

(a) What is d 4 f/dx 2 dydzl 

(b) What is d s f/dx*dydzl 

(c) What is 9 15 //9* 13 3v9z? 

27. Recall from §2.2 that a polynomial in two variables x 
and y is an expression of the form 

d 

p(x, y)=Yl c kix k y', 

k.l=0 

where c« can be any real number for 0 < k, I < d. The 
degree of the term cux k y [ when cu ^ 0 is k + I and 
the degree of the polynomial p is the largest degree 
of any nonzero term of the polynomial (i.e., the largest 
degree of any term for which cu / 0). For example, 
the polynomial 

p(x, y) = 7*y + 2x 2 y 3 - 3x 4 - 5xy 3 + 1 

has five terms of degrees 15, 5, 4, 4, and 0. The de- 
gree of p is therefore 1 5 . (Note: The degree of the zero 
polynomial p(x, y) = 0 is undefined.) 

(a) If p(x, y) = 8jc 7 v 10 — 9x 2 y + 2x, what is the de- 
gree of dp/dx7 dp/dyi d 2 p/dx 2 7 3 2 p/dy 2 ? 
d 2 p/dxdy? 

(b) If p(x, y) = Sx 2 y + 2x 3 y, what is the degree of 
dp/dx? dp/dyl d 2 p/dx 2 7 d 2 p/dy 2 ? d 2 p/dxdy? 

(c) Try to formulate and prove a conjecture relating 
the degree of a polynomial p to the degree of its 
partial derivatives. 

28. The partial differential equation 

3 2 / 3*7 3 2 / 

— - H H = 0 

dx 2 dy 2 dz 2 

is known as Laplace's equation, after Pierre Simon 
de Laplace (1749-1827). Any function / of class C 2 



1 42 Chapter 2 | Differentiation in Several Variables 



that satisfies Laplace's equation is called a harmonic 
function. 3 

(a) Is f(x, y, z) = x 2 + y 2 — 2z 2 harmonic? What 
about f(x, y, z) = x 2 - y 2 + z 2 1 

(b) We may generalize Laplace's equation to functions 
of n variables as 

3 2 / 3 2 / 9 2 / „ 

— y + — T + ■■■+— V=0. 

ox{ 9xj 9jc„ 

Give an example of a harmonic function of n vari- 
ables, and verify that your example is correct. 

29. The three-dimensional heat equation is the partial dif- 
ferential equation 

/d 2 T d 2 T d 2 T\ _ dT 
\~d~x 2 + ~d~y 2 + Ik 2 ) ~ ~dt' 

where k is a positive constant. It models the tempera- 
ture T(x, y, z, t) at the point (x, y, z) and time t of a 
body in space. 

(a) We examine a simplified version of the heat equa- 
tion. Consider a straight wire "coordinatized" by 
x. Then the temperature T(x, t) at time t and po- 
sition x along the wire is modeled by the one- 
dimensional heat equation 

d 2 T _ dT 
Jx 2 ~ ~dt' 

Show that the function T(x, t) = e~ kt cosjc satis- 
fies this equation. Note that if t is held constant at 
value ?o, then T(x, to) shows how the temperature 
varies along the wire at time to. Graph the curves 
z = T(x, to) for to = 0,1, 10, and use them to un- 
derstand the graph of the surface z = T(x,t) for 
t > 0. Explain what happens to the temperature of 
the wire after a long period of time. 

(b) Show that T(x, y, t) = e~ k '(cosx + cos y) satis- 
fies the two-dimensional heat equation 

id 2 T d 2 T\ _ dT 
[jx* + Jy 2 ) = It' 



Graph the surfaces given by z = T(x, y, to), where 
to = 0, 1, 10. If we view the function T(x, y, t) as 
modeling the temperature at points (x, y) of a flat 
plate at time t, then describe what happens to the 
temperature of the plate after a long period of time. 

(c) Now show that T(x, y, z, t) = e~ kt (cosx + 
cos y + cos z) satisfies the three-dimensional heat 
equation. 

30. Let 

[-^(4x4) 30^(0,0) 
/( je>y )=J \x 2 + y 2 J 

[ 0 if(x,y) = (0,0) 

(a) Find f x (x,y) and /,(* , y) for (x, y) ? (0, 0). (You 
will find a computer algebra system helpful.) 

(b) Either by hand (using limits) or by means of 
part (a), find the partial derivatives / r (0, y) and 
/,(x,0). 

(c) Find the values of f xy (0, 0) and /y.,-(0, 0). Recon- 
cile your answer with Theorem 4.3. 

A surface that has the least surface area among all surfaces 
with a given boundary is called a minimal surface. Soap bub- 
bles are naturally occurring examples of minimal surfaces. It 
is a fact that minimal surfaces having equations of the form 
z = fix, y) (where f is of class C 2 ) satisfy the partial differ- 
ential equation 

(1 + l]) Z XX + (1 + Zl) Zyy = 2 Zx ZyZ xy . (6) 

Exercises 31—33 concern minimal surfaces and equation (6). 

31 . Show that a plane is a minimal surface. 

32. Scherk's surface is given by the equation e z cos y = 
cosx. 

(a) Use a computer to graph a portion of this surface. 

(b) Verify that Scherk's surface is a minimal surface. 

33. One way to describe the surface known as the helicoid 
is by the equation x = y tan z. 

(a) Use a computer to graph a portion of this surface. 

(b) Verify that the helicoid is a minimal surface. 



2.5 The Chain Rule 

Among the various properties that the derivative satisfies, one that stands alone 
in both its usefulness and its subtlety is the derivative's behavior with respect 
to composition of functions. This behavior is described by a formula known as 



3 Laplace did fundamental and far-reaching work in both mathematical physics and probability theory. 
Laplace's equation and harmonic functions are part of the field of potential theory, a subject that Laplace 
can be credited as having developed. Potential theory has applications to such areas as gravitation, elec- 
tricity and magnetism, and fluid mechanics, to name a few. 



2.5 | The Chain Rule 143 



the chain rule. In this section, we review the chain rule of one- variable calculus 
and see how it generalizes to the cases of scalar- and vector-valued functions of 
several variables. 

The Chain Rule for Functions of One Variable: A Review 

We begin with a typical example of the use of the chain rule from single-variable 
calculus. 

EXAMPLE 1 Let f(x) = sinx and x(t) = t 3 + t. We may then construct the 
composite function /(x(r)) = sin(f 3 + t). The chain rule tells us how to find the 
derivative of/oi with respect to t : 

(/ o jc)'(r) = -^-(sin(7 3 + 0) = (cos(? 3 + t))(3t 2 + 1). 
at 

Since x = r 3 + t, we have 

(/ o x)'(t) = isinjc) ■ ^{t 1 + 0 = f{x) ■ x'(t). ♦ 
ax at 

In general, suppose X and T are open subsets of R and /:XcR^ R 
and x : T C R — >• R are functions defined so that the composite function 
/ o x : T -> R makes sense. (See Figure 2.57.) In particular, this means that the 
range of the function x must be contained in X, the domain of /. The key result 
is the following: 



/ 



-i ) ( — h 



X 

R R R 

Figure 2.57 The range of the function x must be contained in the domain X of / in 
order for the composite / o x to be defined. 



THEOREM 5.1 (The chain rule in one variable) Under the preceding as- 
sumptions, if x is differentiable at to 6 Tand /isdifferentiableat.To = x(to) 6 X, 
then the composite / o x is differentiable at to and moreover, 

(/ o x)'(to) = f\xo)x'(to). (1) 

A more common way to write the chain rule formula in Theorem 5.1 is 
df df dx 

Mt 0 ) = Mx 0 ) —(to). (2) 
dt dx dt 

Although equation (2) is most useful in practice, it does represent an unfor- 
tunate abuse of notation in that the symbol / is used to denote both a function 
of x and one of t. It would be more appropriate to define a new function y by 
y(t) = (f o x)(t) sothatdy/df = (df/dx)(dx/dt). But our original abuse of no- 
tation is actually a convenient one, since it avoids the awkwardness of having too 
many variable names appearing in a single discussion. In the name of simplicity, 
we will therefore continue to commit such abuses and urge you to do likewise. 

The formulas in equations (1) and (2) are so simple that little more needs 
to be said. We elaborate, nonetheless, because this will prove helpful when we 



1 44 Chapter 2 | Differentiation in Several Variables 



generalize to the case of several variables. The chain rule tells us the following: 
To understand how / depends on t, we must know how / depends on the "in- 
termediate variable" x and how this intermediate variable depends on the "final" 
independent variable t. The diagram in Figure 2.58 traces the hierarchy of the 
variable dependences. The "paths" indicate the derivatives involved in the chain 
rule formula. 




Figure 2.58 The chain rule for functions of a single variable. 




Figure 2.59 The composite function fox. 



The Chain Rule in Several Variables 

Now let's go a step further and assume /:XcR 2 ^RisaC' function of 
two variables and x: T c R -> R 2 is a differentiable vector-valued function 
of a single variable. If the range of x is contained in X, then the composite 
/ox:TcR^Ris defined. (See Figure 2.59.) It's good to think of x as 
describing a parametrized curve in R 2 and / as a sort of "temperature func- 
tion" on X. The composite / o x is then nothing more than the restriction of / to 
the curve (i.e., the function that measures the temperature along just the curve). 
The question is, how does / depend on t ? We claim the following: 



PROPOSITION 5.2 Suppose x: T C R -> R 2 is differentiable at t 0 e T, and 
/: X C R 2 —> R is differentiable at xo = x(?o) = (xq, yo) £ X, where T and X 
are open in R and R 2 , respectively, and range x is contained in X. If, in addition, 
/ is of class C 1 , then fox:T^ R is differentiable at to and 

df df dx df dy 

dt dx dt dy dt 



Before we prove Proposition 5.2, some remarks are in order. First, notice 
the mixture of ordinary and partial derivatives appearing in the formula for the 



2.5 | The Chain Rule 145 




derivative. These terms make sense if we contruct an appropriate "variable hier- 
archy" diagram, as shown in Figure 2.60. At the intermediate level, / depends 
on two variables, x and y (or, equivalently, on the vector variable x = (x , y)), 
so partial derivatives are in order. On the final or composite level, / depends on 
just a single independent variable t and, hence, the use of the ordinary derivative 
df /dt is warranted. Second, the formula in Proposition 5.2 is a generalization of 
equation (2): A product term appears for each of the two intermediate variables. 




Figure 2.61 The graph of the 
function x of Example 2. 



EXAMPLE 2 Suppose f(x, y) = (x + y 2 )/(2x 2 + 1) is a temperature func- 
tion on R 2 and x(?) = (2t, t + 1). The function x gives parametric equations 
for a line. (See Figure 2.61.) Then 



(/ox)(0 = /(x(0) 



2t + (t + iy t 2 + At + i 



8r 2 + 1 8? 2 + 1 

is the temperature function along the line, and we have 

df A- lAt - 32t 2 



dt 



(St 2 + l) 2 



by the quotient rule. Thus, all the hypotheses of Proposition 5.2 are satisfied and 
so the derivative formula must hold. Indeed, we have 



and 



Therefore, 



dx 
df 

dy 



At) 



\-2x 2 



Axy 2 



(2x 2 + l) 2 
2y 



2x 2 + r 

dx dy 
dt ' dt 



(2, 1). 



df dx df dy 
dx dt dy dt 



1 - 2x 2 - Axy 2 
(2x 2 + l) 2 

2(1 - St 2 - 8t(t 



2 + 



2v 



2jc 2 + 1 



(8f 2 + l) 2 



}f) + 2(t 



1) 



8f 2 + 1 



1 46 Chapter 2 | Differentiation in Several Variables 



after substitution of 2t for x and t + 1 for y. Hence, 

df dx df dy _ 2(2 -It- \6t 2 ) 
dx~dt + ~dy~dt ~ (8? 2 + l) 2 ' 
which checks with our previous result for df/dt. ♦ 

Proof of Proposition 5.2 Denote the composite function / o x by z- We want to 
establish a formula for dz/dt at to- Since z is just a scalar- valued function of one 
variable, differentiability and the existence of the derivative mean the same thing. 
Thus, we consider 

dz z(t)-z(t 0 ) 

—(to) = km — - — , 

dt t^t 0 t — to 

and see if this limit exists. We have 

dz,, f(x(t), y(t)) - f(x(t 0 ), y(t 0 )) 

—(to) = bm . 

dt f-»-«o t — to 

The first step is to rewrite the numerator of the limit expression by subtracting 
and adding f(x 0l y) and to apply a modicum of algebra. Thus, 

dz r^ v f( x ' - /(*o» y) + f(xo, y) - f(x 0 , y 0 ) 

—(to) = bm 

dt t->t Q t — to 

,. fix, y) - f(x 0 , y) f(xo, y) - f(x 0 , y 0 ) 
= hm h bm . 

t-*h t — to t-*t<, t — to 

(Remember that x(to) = xo = (xo, yo)-) Now, for the main innovation of the proof. 
We apply the mean value theorem to the partial functions of /. This tells us that 
there must be a number c between x 0 and x and another number d between y 0 
and y such that 



and 



Thus, 



f(x, y) - f(x 0 , y) = f x (c, y)(x - x 0 ) 
f(x 0 , y) - f(x 0 , yo) = f y (x 0 , d)(y - y 0 ). 



dz x — xq y — yo 

—(to) = hm f x (c, y)- — — + hm f y (x 0 , d)- — — 
dt t->t 0 t — to ■ t — to 

x(t)-x(to) ,,y(t)-y(to) 

= hm f x (c, y) — h hm f y (x 0 , d) — 

t-*k t - 1 0 t^-to t - to 

dx dy 

= fx(XQ, yo)-J-(to) + fy(XQ, yo)-T-(t 0 ), 

dt dt 
by the definition of the derivatives 

dx dy 
-(to) and -(t 0 ) 

and the fact that f x (c, y) and f y (xo, d) must approach f x (xo, yo) and f y (xo, yo), 
respectively, as t approaches to, by continuity of the partials. (Recall that / was 
assumed to be of class C 1 . ) This completes the proof. ■ 

Proposition 5.2 and its proof are easy to generalize to the case where / 
is a function of n variables (i.e., /:XC R" —> R) and x : T C R — >• R". The 



2.5 | The Chain Rule 147 



appropriate chain rule formula in this case is 

df , , df , . dxi , df , ,dx2, s 

= -^(x 0 )— ^ 0 ) + -^(x 0 )— i(r 0 ) + 
at ax\ at 0x2 dt 



df ,dx„ 

+ w~ (xo)-r-(fo)- 
3x„ a/ 



(3) 



Note that the right side of equation (3) can also be written by using matrix notation 
so that 



dj_ 
dt 



Co) = 



9/ 
3xi 



(xo) 



Thus, we have shown 



3/ 
3x 2 



(xo) 



dx„ 



(xo) 



~dl 



dt 



(to) 
(to) 

(to) 



df 
dt 



(t 0 ) = Df(x 0 )Dx(t 0 ) = V/(x 0 ) -x'(/ 0 ), 



(4) 



where we use x'(fo) as a notational alternative to Dx(fo)- The version of the chain 
rule given in formula (4) is particularly important and will be used a number of 
times in our subsequent work. 



Let us consider further instances of composition of functions of many 
variables. For example, suppose X is open in R 3 , T is open in R 2 , and 
/:XcR 3 ^R and x: T c R 2 R 3 are such that the range of x is contained 
in X. Then the composite / os:rcR 2 ->R can be formed as shown in 
Figure 2.62. Note that the range of x, that is, x(7), is just a surface in R 3 , so 
fox can be thought of as an appropriate "temperature function" restricted to this 
surface. If we use x = (x, y, z) to denote the vector variable in R 3 and t = (s,t) 
for the vector variable in R 2 , then we can write a plausible chain rule formula 
from an appropriate variable hierarchy diagram. (See Figure 2.63.) Thus, it is 



X 




1 48 Chapter 2 | Differentiation in Several Variables 




Figure 2.63 The chain rule for / o x, where /:XCR ! ^R and 
\:T c R 2 ->- R 3 . 



reasonable to expect that the following formulas hold: 

df = dfdx + dfdy dfdz 

ds dx ds dy ds dz ds 

and (5) 

df _ df dx df dy df dz 
~dl ~ dx~dl + ~dyJt + JzJt' 

(Again, we abuse notation by writing both df/ds, df/dt and df/dx, df/dy, 
df/dz.) Indeed, when / is a function of x, y, and z of class C 1 , formula (3) 
with n = 3 applies once we realize that dx/ds, dx/dt, etc., represent ordinary 
differentiation of the partial functions in s or t. 

EXAMPLE 3 Suppose 

f(x, y, z) = x 2 + y 2 + z 2 and x(s, t) = (s cos t, e st , s 2 - t 2 ). 
Then h(s, t) = f o x(s, t) = s 2 cos 2 1 + e 2st + (s 2 - t 2 ) 2 , so that 

dh d(f ox) ^ 2 ~ , , 7x 

= 2* cos 2 f + 2fe 2sr + 4s(s 2 - t 2 ) 



dt dt 
We also have 



ds ds 

= -2s 1 cos t sin t + 2se Ist - 4t(s z - t l ). 



dh d(fox) 2 ,, 2 



and 



9/ - 3/ - 3/ - 
— = 2x, — = 2y, — = 2z 
dx dy dz 



dx dx 

— = cosr, — = — jsin/, 
ds dt 

t st d y st 

— = te , — = se , 

ds dt 

3z dz 

— =2s, — = -It. 
ds dt 



2.5 | The Chain Rule 149 



Hence, we compute 

df = d(f o x) = df dx | df dy | df dz 
ds ds dx ds dy ds dz ds 

= 2x(cos 0 + 2y(te st ) + 2z(2s) 

= 2s cos r(cos 0 + 2e st (te st ) + 2{s 2 - t 2 )(2s) 

= 2s cos 2 t + 2te 2s1 + 4s(s 2 - t 2 ), 

just as we saw earlier. We leave it to you to use the chain rule to calculate df/dt 
in a similar manner. ♦ 

Of course, there is no need for us to stop here. Suppose we have an open set 
X in R'", an open set T in R", and functions /: X —> R and x: T — »• R'" such 
that A = / o x: I ^> R can be defined. If / is of class C 1 and x is differentiate, 
then, from the previous remarks, h must also be differentiate and, moreover, 

dh df dxi df 3x2 3/ dx m 

dtj dx\ dtj dx2 dtj dx m dtj 



= E 



7 = 1,2, 



df dx k 
k=l dx k dtj 

Since the component functions of a vector- valued function are just scalar- valued 
functions, we can say even more. Suppose f: X c R m -> RP and x: T c R" 
R" ! are such that h = f o x: T c R" — >• R'' can be defined. (As always, we assume 
that X is open in R m and T is open in R" .) See Figure 2.64 for a representation of 
the situation. If f is of class C 1 and x is differentiable, then the composite h = f o x 
is differentiable and the following general formula holds: 



dhi 
3^ 



df dx k 
dxk dt J 



1,2, 



7 = 1,2, 



, n. 



(6) 



The plausibility of formula (6) is immediate, given the variable hierarchy diagram 
shown in Figure 2.65. 




X 



x(T) 



R" R m Rp 

Figure 2.64 The composite fo x where f : X C R"' ^ R'' and x: T C R" R m . 

Now comes the real "magic." Recall that if A is a p x m matrix and B is an 
m x n matrix, then the product matrix C = AB is defined and is a /? x « matrix. 
Moreover, the i/th entry of C is given by 



kj- 



k=l 



Chapter 2 | Differentiation in Several Variables 




If we recall that the ijth entry of the matrix Dh(t) is dhi/dtj, and similarly for 
Z)f(x) and Dx(t), then we see that formula (6) expresses nothing more than the 
following equation of matrices: 

Dh(t) = D(f o x)(t) = Df(x)Dx(t). (7) 

The similarity between formulas (7) and (1) is striking. One of the reasons 
(perhaps the principal reason) for defining matrix multiplication as we have is 
precisely so that the chain rule in several variables can have the elegant appearance 
that it has in formula (7). 

EXAMPLE 4 Suppose f:R 3 -> R 2 is given by f(xi,x 2 ,x 3 ) = 
(jci — x 2 , xix 2 x 3 ) and x: R 2 -> R 3 is given by x(*i, t 2 ) = (ht2, t 2 
f o x: R 2 -> R 2 is given by (f o x)(*i , t 2 ) = (tit 2 - tf, f 3 f 2 3 ), so that 



ls 4). Then 



D(fox)(t) 

On the other hand 

Df( X ) : 

so that the product matrix is 
Df(x)Dx(t) 



t 2 - 2fi 
3t 2 4 



h 

3ty 2 



1 -1 0 

X 2 Xj, X\Xt, X\X 2 



h ~ 2*i 



and Dx(t) 



h *i 
2*! 0 
0 2*2 



X 2 X-x,t 2 + 2x 1X3*1 X2*3fi + 2x\X 2 t 2 



*2 

t 2 t 3 

'1 '2 



2*1 
2f 2 



,3 

1 '2 



2t\t\ 



after substituting forxi, x 2 , andx 3 . Thus, D(f o x)(t) = Df(x)Dx(t), as expected. 

Alternatively, we may use the variable hierarchy diagram shown in Figure 2.66 
and compute any individual partial derivative we may desire. For example, 

dfl _ df^dx^ dfidx^ df 2 9x 3 
3*i 9xi 3*i 3x2 3?i 3x3 3*i 



2.5 | The Chain Rule 151 




Figure 2.66 The variable hierarchy diagram for Example 4. 

by formula (6). Then by abuse of notation, 

3/ 2 



= (*2*3)(fc) + (*l*3)(2fi) + (XIX 2 )(0) 



= (t 2 t 2 2 )(t 2 ) + ( h t 2 )(t 2 2 )(2 h ) 



which is indeed the (2,1) entry of the matrix product. ♦ 

At last we state the most general version of the chain rule from a technical 
standpoint; a proof may be found in the addendum to this section. 



THEOREM 5.3 (The chain rule) Suppose X c R" ! and T c R" are open and 
f: X — > K p and x: T — »• R'" are defined so that range x c X. If x is differen- 
tiable at t 0 6 T and f is differentiable at x 0 = x(t 0 ), then the composite f o x is 
differentiable at to, and we have 



D(fox)(t 0 )= Df(x 0 )Dx(t 0 ). 



The advantage of Theorem 5.3 over the earlier versions of the chain rule we 
have been discussing is that it requires f only to be differentiable at the point in 
question, not to be of class C 1 . Note that, of course, Theorem 5.3 includes all 
the special cases of the chain rule we have previously discussed. In particular, 
Theorem 5.3 includes the important case of formula (4). 

EXAMPLE 5 Let f: R 2 -> R 2 be defined by f(x, y) = (x - 2y + 7, 3xy 2 ). 
Suppose that g: R 3 -> R 2 is differentiable at (0, 0, 0) and we know that 
g(0, 0, 0) = (-2, 1) and 

2 4 5 " 
Dg(0,0,0)= _i o 1 ■ 

We use this information to determine D(f o g)(0, 0, 0). 

First, note that Theorem 5.3 tells us that fog must be differentiable at (0, 0, 0) 
and, second, that 



D(f o g)(0, 0, 0) = Df (g0, 0, 0)) Dg(0, 0, 0) = Df(-2, l)Dg(0, 0, 0). 



Chapter 2 | Differentiation in Several Variables 



Since we know f completely, it is easy to compute that 
Df(x,y) = 

Thus, 

D(fog)(0,0,0)= 3 

We remark that we needed the full strength of Theorem 5.3, as we do not know 
anything about the differentiability of g other than at the point (0, 0, 0). ♦ 



1 -2 

3y 2 6xy 



so that Df(-2, 1) 



-2 
■12 



2 4 5 
-1 0 1 



4 3 
12 3 



EXAMPLE 6 (Polar/rectangular conversions) Recall that in §1.7 we pro- 
vided the basic equations relating polar and rectangular coordinates: 

x = r cos 9 
y = r sin 9 

Now suppose you have an equation defining a quantity w as a function of x and 
y; that is, 

w = f{x, y). 

Then, of course, w may just as well be regarded as a function of r and 9 by 
susbtituting r cos 9 for x and r sin 9 for y • That is, 



g(r, 0) = f(x(r, 9), y(r, 9)). 



Our question is as follows: Assuming all functions involved are differentiable, 
how are the partial derivatives dw/dr, dw/d9 related to dw/dx, dw/dy? 

In the situation just described, we have w = g(r, 9) = (/ o x)(r, 9), so that 
the chain rule implies 



Dg(r,9) = Df(x,y)Dx(r,9). 



Therefore, 



dg 
8r 



9g 
36 



dx 



df 

dx 



df 

dy 

df 
dy 



dx 


dx 




W 


dy 


dy 


dr 


d9 _ 


cos 9 


—r sin 


sin 9 


r cos 



By extracting entries, we see that the various partial derivatives of w are related 
by the following formulas: 



3w 
"97 
dw 
~d~9 



cost 



dw dw 

h sm9 — 

dx dy 



dw dw 

-r smw h r zos9 — 

dx dy 



(8) 



The significance of (8) is that it provides us with a relation of differential 
operators: 



2.5 | The Chain Rule 153 



a a 

— = cost? — 
dr dx 



sinO 



dy 



d a a 

— = —r sin 6 h r cos 6 — 

dd dx dy 



(9) 



The appropriate interpretation for (9) is the following: Differentiation with 
respect to the polar coordinate r is the same as a certain combination of differen- 
tiation with respect to both Cartesian coordinates x and y (namely, the combina- 
tion cos 0 d/dx + sm6 d/dy). A similar comment applies to differentiation with 
respect to the polar coordinate 0 . Note that, when r ^ 0, we can solve algebraically 
for d/dx and d/dy in (9), obtaining 



a 

— = cosf 
dx 




sin 6 


d 


dr 


/• 


dQ 


a 

— = sinfi 

dy 


a 

Yr + 


cos 6 


a 


r 


d9 



(10) 



We will have occasion to use the relations in (9) and (10), and the method of 
their derivation, later in this text. ♦ 

Addendum: Proof of Theorem 5.3 



We begin by noting that the derivative matrices £>f(x 0 ) and Dx(t 0 ) both exist 
because f is assumed to be differentiable at x 0 and x is assumed to be differentiable 
at to. Thus, the product matrix Df(xo)£>x(to) exists. We need to show that the limit 
in Definition 3.8 is satisfied by this product matrix, that is, that 

||(fox)(t)-[(fox)(t 0 ) + Df(xo)Dx(t 0 )(t-to)]|| 

hm = 0. (11) 

||t -toll 

In view of the uniqueness of the derivative matrix, it then automatically follows 
that f o x is differentiable at to and that Z)f(xo)£>x(to) = D(f o x)(to). Thus, we 
entirely concern ourselves with establishing the limit (11) above. 
Consider the numerator of (1 1). First, we rewrite 

(f o x)(t) - [(f ox)(to) + Df(x 0 )Dx(t 0 )(t - t 0 )] 

= (f o x)(t) - (f o x)(t 0 ) - £>f(x 0 )(x(t) - x(f 0 )) 

+ Df(x 0 )(x(t) - x(f 0 )) - Df(xo)Dx(t 0 )(t - t 0 ). 

Then we use the triangle inequality: 

|| (f o x)(t) - [(f o x)(t 0 ) + £>f(x 0 )Dx(to)(t - t 0 )] || 

< || (f o x)(t) - (f o x)(t 0 ) - Df(x 0 )(x(t) - x(t 0 ))\\ 

+ ||£»f(x 0 )(x(t) - x(f 0 )) - Df(x 0 )Dx(t 0 )(t - t 0 ) || 
= || (f o x)(t) - (f o x)(t 0 ) - Df(x 0 )(x(t) - x( ?0 ))|| 

+ ||Df(xo) [(x(t) - x(f 0 )) - Dx(t 0 )(t - t 0 )] || . 



1 54 Chapter 2 | Differentiation in Several Variables 



By inequality (9) in the proof of Theorem 3.9, there is a constant K such that, for 
any vector h e R", ||Df(x 0 )h|| < A"||h||. Thus, 

|| (fo x)(t) - (f o x)(t 0 ) - £>f(x 0 )Dx(t 0 )(t - t 0 ) || 

< || (f o x)(t) - (f o x)(t 0 ) - Df(x 0 )(x(t) - x(t 0 ))\\ (12) 
+ K||x(t)-x(fo)- J Dx(t 0 Xt-to)||. 

To establish the limit (11) formally, we must show that given any e > 0, we 
may find a S > 0 such that if 0 < ||t — toll < 8, then 

|| (f o x)(t) - [(f o x)(t 0 ) + £>f(xo)Z)x(to)(t - t 0 )] || <f 

lit -toll 

Consider the first term of the right side of (12). Using the differentiability of x at 
t 0 and inequality (1 1) in the proof of Theorem 3.9, we can find some <5 0 > 0 and 
a constant A"o such that if 0 < ||t — toll < So, then 

||x(t) - x(t 0 )|| < K 0 ||t -to||. 

By the differentiability of f at xo, given any e\ > 0, we may find some Si > 0 
such that if 0 < ||x — Xo|| < <$i, then 

||f(x)-[f(xo) + Df(x 0 )(x-xo)] || 

< ei- 



l|x-xo|| 

Setei = e/(2A: 0 ). Withx = x(t),x 0 = x(t 0 ), we have that if both 0 < ||t- t 0 || < 
So and 0 < ||t — t 0 || < 8\/Kq, then 

||x(t) - x(t 0 )|| < Ko lit -toll <<5i. 

Hence, 

||(f o x)(t) - (f o x)(t 0 ) - Z)f(x 0 )(x(t) - x(t 0 ))|| < ei||x(t) - x(t 0 )|| 

< ei^ollt- toll = . (13) 

Now look at the second term of the right side of (12). Since x is differentiable 
at t 0 , given any €2 > 0, we may find some S 2 > 0 such that if 0 < ||t — t 0 1| < S 2 , 
then 

||x(t) - [x(t 0 ) + £>x(t 0 )(t - t 0 )] || 
lit -toll 

Set e 2 = e/(2K). Then, for 0 < ||t - t 0 || < S 2 , we have 

||x(t)-[x(to) + Dx(to)(t-t 0 )]|| < 2^llt-t 0 ||. (14) 

Finally, let S be the smallest of <5o, <5i/A"o, and 82- Then, for 0 < ||t — to || < 8, we 
have that both the inequalities (13) and (14) hold and thus (12) becomes 

|| (fo x)(t) - (f o x)(t 0 ) - Df(x 0 )£»x(t 0 )(t - t 0 ) || 

< € -\\t - toll +^(^11*- toll) 

= e||t- toll- 
Hence, 

||(f o x)(t) - (f o x)(t 0 ) - £)f(xo)Dx(to)(t - t 0 ) " 
lit -toll 

as desired. 



2.5 I Exercises 155 



2.5 Exercises 



1. If f(x, y, z) = x 2 — y 3 + xyz, and x = 6t + 7, y = 
sin2r, z = t 2 , verify the chain rule by finding df /dt 
in two different ways. 



months from now will be 



71 1 



s 2 + t 2 , 



find 



2. If f(x, y) = sin(;ty) and x = s + t, y 
df/ds and df/dt in two ways: 

(a) by substitution. 

(b) by means of the chain rule. 

3. Suppose that a bird flies along the helical curve 
x = 2 cos t, y = 2 sin t,z = 3f . The bird suddenly en- 
counters a weather front so that the barometric pres- 
sure is varying rather wildly from point to point as 
P(x, y, z) = 6x 2 z/y atm. 

(a) Use the chain rule to determine how the pressure 
is changing at t = rc/4 min. 

(b) Check your result in part (a) by direct substitution. 

(c) What is the approximate pressure at t = rc/4 + 
0.01 min? 



4. Suppose that z 



+ y , where x = st and y is a 



function ofs and t . Suppose further that when (s, t) = 

dz 

(2, 1), dy/dt = 0. Determine —(2, 1). 

5. You are the proud new owner of an Acme Deluxe Bread 
Kneading Machine, which you are using for the first 
time today. Suppose that at noon the dimensions of your 
(nearly rectangular) loaf of bread dough are L = 7 in 
(length), W = 5 in (width), and H = 4 in (height). At 
that time, you place the loaf in the machine for knead- 
ing and the machine begins by stretching the loaf's 
length at an initial rate of 0.75 in/min, punching down 
the loaf's height at a rate of 1 in/min, and increasing 
the loaf's width at a rate of 0.5 in/min. What is the rate 
of change of the volume of the loaf when the machine 
starts? Is the dough increasing or decreasing in size at 
that moment? 

6. A rectangular stick of butter is placed in the microwave 
oven to melt. When the butter's length is 6 in and its 
square cross section measures 1 .5 in on a side, its length 
is decreasing at a rate of 0.25 in/min and its cross- 
sectional edge is decreasing at a rate of 0.125 in/min. 
How fast is the butter melting (i.e., at what rate is the 
solid volume of butter turning to liquid) at that instant? 

7. Suppose that the following function is used to model 
the monthly demand for bicycles: 

P(x, y) = 200 + 20V0.1* + 10 - lltfy. 

In this formula, x represents the price (in dollars 
per gallon) of automobile gasoline and y repre- 
sents the selling price (in dollars) of each bicycle. 
Furthermore, suppose that the price of gasoline t 



x = 1 + O.lf — cos ■ 
b 

and the price of each bicycle will be 

TCt 

y = 200 + 2t sin — . 

6 

At what rate will the monthly demand for bicycles be 
changing six months from now? 

8. The Centers for Disease Control and Prevention pro- 
vides information on the body mass index (BMI) 
to give a more meaningful assessment of a person's 
weight. The BMI is given by the formula 



BMI : 



10,000u; 



where w is an individual's mass in kilograms and h 
the person's height in centimeters. While monitoring a 
child's growth, you estimate that at the time he turned 
1 0 years old, his height showed a growth rate of 0.6 cm 
per month. At the same time, his mass showed a growth 
rate of 0.4 kg per month. Suppose that he was 140 cm 
tall and weighed 33 kg on his tenth birthday. 

(a) At what rate is his BMI changing on his tenth 
birthday? 

(b) The BMI of a typical 10-year-old male increases 
at an average rate of 0.04 BMI points per month. 
Should you be concerned about the child's weight 
gain? 

9. A cement mixer is pouring concrete in a conical pile. 
At the time when the height and base radius of the 
concrete cone are, respectively, 30 cm and 12 cm, the 
rate at which the height is increasing is 1 cm/min and 
the rate at which the volume of cement in the pile is 
increasing is 320 cm 3 /min. At that moment, how fast 
is the radius of the cone changing? 

1 0. A clarinetist is playing the glissando at the beginning of 
Rhapsody in Blue, while Hermione (who arrived late) 
is walking toward her seat. If the (changing) frequency 
of the note is / and Hermione is moving toward the 
clarinetist at speed v, then she actually hears the fre- 
quency (j> given by 



c + v 



f, 



where c is the (constant) speed of sound in air, about 
330 m/sec. At this particular moment, the frequency is 
/ = 440 Hz and is increasing at a rate of 100 Hz per 
second. At that same moment, Hermione is moving 
toward the clarinetist at 4 m/sec and decelerating at 



Chapter 2 | Differentiation in Several Variables 



2 m/sec 2 . What is the perceived frequency cf> she 
hears at that moment? How fast is it changing? Does 
Hermione hear the clarinet's note becoming higher or 
lower? 

11. Suppose z = f(x,y) has continuous partial deriva- 
tives. Let x = e'cos8, y = e'smd. Show that 
then 



+ 



-2r 



+ 



dz. 



X z 



1 8. Suppose that ui = g I — , — I is a differentiable func- 

\y yJ 

tion of u = x/y and v = z/y. Show then that 



dw dw dw 

x ir + yir + z ir= 0 - 

ox ay dz 



In Exercises 19-27, calculate D(io g) in two ways: (a) by first 
evaluating fog and (b) by using the chain rule and the deriva- 
tive matrices Df and Dg. 



12. Suppose that z = f(x, y) has continuous partial 
derivatives. Let x = 2uv and y = u 2 + v 2 . Show that 
then 



dz dz 



2x 



du dv 
13. If w = g(u 2 



+ 



, dz dz 

+ 4v . 

dx 3y 



u 2 ) has continuous partial 



derivatives with respect to x = u 2 — v 2 and y = v 2 — 
u 2 , show that 

dw dw 

v h u — =0. 

du dv 

14. Suppose that z = f(x + y, x — y) has continuous par- 
tial derivatives with respect to u = x + y and v = 
x — y. Show that 



dz 


dz 


(du) 


(dz 


dx 


dy " 




\dv 



15. If w = f 



xy 



xy 



x 2 + y 



is a differentiable function of 



- , show that 



x 2 + y 2 

dw dw 

x h y = 0. 

dx dy 

(x 2 -y 2 \ 

16. If w = f —z r- is a differentiable function of 

\x 2 + y 2 J 

x 2 - y 2 
u = — , show that then 



x l + y 1 



dw dw 

x h y = 0. 

dx dy 



I y — x z — x \ 
1 7. Suppose w = f [ , ) is a differentiable 

V xy xz 

y — x z — x 
function of u = and v = . Show then that 



xy 



xz 



r. dw i dw i dw 

x ir + y IT + z IT = °- 

dx dy dz 



19. f(x) = (3x 5 , e 2x ), g(s, t) = s- It 

20. f(x) = (x 2 , cosSx, Inx), g(s, t,u) = s + t 2 + i 

21. f(x, y) = ye x , g(s, t) = (s - t, s + t) 



22. f(x, y) 

23. f(x, y) = 



3y 2 ,g(s,t) = (st,s + t 2 ) 



(xv-^ + v 3 ),g(,,r)=(^ 2 r) 



y x 
x' y 

24. f(x, y, z) = (x 2 y + y 2 z, xyz, e z ), 
g(t) = (f - 2, 3f + 7, f 3 ) 

25. f(x, y) = (xy 2 , x 2 y,x 3 + y 3 ), g(f) = (sinf, e') 

26. f(x, y) = (x 2 - y, y/x, e y ), g(s, t, u) = (s + It + 
3m, stu) 

27. f(x, y, z) = (x + y + z, x 3 - e vz ), 
g(s, t, u) = (st, tu, su) 

28. Let g: R 3 — >• R 2 be a differentiable function 
such that g(l,-l,3) = (2, 5) and Dg(l,-1,3) = 

1 -1 0 



0 7 



Suppose that f: R — > R is de- 



fined by f(x, v) = (2xy, 3x — y + 5). What is 
D(fog)(l,-l,3)? 

29. Let g:R 2 ^R 2 and f:R 2 ^R 2 be differentiable 
functions such that g(0, 0) = (1, 2), g(l,2) = 
(3,5), f(0, 0) = (3, 5), f(4, 1) = (1,2), Z)g(0,0) = 



1 0 
-1 4 



1 1 

3 5 



Dg{\,2): 

Df(4, 1) = 



2 3 
5 7 



£>f(3, 5) = 



1 2 
1 3 



(a) Calculate D(fog)( 1,2). 

(b) Calculate D(gof)(4, 1). 

30. Let z = f(x,y), where / has continuous partial 
derivatives. If we make the standard polar/rectangular 
substitution x = r cos 0, y = r sin 9, show that 



+ 



+ 



1 



.30. 



2.5 I Exercises 157 



31. (a) Use the methods of Example 6 and formula (10) 
in this section to determine d 2 /dx 2 and d 2 /dy 2 
in terms of the polar partial differential operators 
d 2 /dr 2 , d 2 /dd 2 , d 2 /dr d6, d/dr, and d/dd. (Hint: 
You will need to use the product rule.) 

(b) Use part (a) to show that the Laplacian operator 

d 2 /dx 2 + d 2 /dy 2 is given in polar coordinates by 
the formula 



il il 

Sx 2 + dy 2 



dr 2 



+ 



1 3 
r dr 



+ 



l 

3<9 2 ' 



32. Show that the Laplacian operator d 2 /dx 2 + d 2 /dy 2 + 
d 2 /dz 2 in three dimensions is given in cylindrical co- 
ordinates by the formula 



3 2 d 2 d 2 
dx 2 By 2 dz 2 



d 2 13 1 



■dr- 



r dr 



+ 



r 2 d6 2 dz 



33. In this problem, you will determine the formula for the 
Laplacian operator in spherical coordinates. 

(a) First, note that the cylindrical/spherical conver- 
sions given by formula (6) of §1.7 express the 
cylindrical coordinates z and r in terms of the 
spherical coordinates p and <p by equations of pre- 
cisely the same form as those that express x and 
y in terms of the polar coordinates r and 9. Use 
this fact to write 3/3r in terms of 3/3p and d/d<p. 
(Also see formula (10) of this section.) 

(b) Use the ideas and result of part (a) to establish the 
following formula: 

3 2 3 2 3 2 
d^ 2 + dy~ 2 + dz 2 



+ 



1 3 2 



3p 2 p 2 dip 2 



+ 



p z sin" cp 



so 2 



2 3 cot go 

p dp p 2 



dcp 



34. Suppose that y is defined implicitly as a function y(x) 
by an equation of the form 

F(x, y) = 0. 

(For example, the equation x 3 — y 2 = 0 defines y as 
two functions of x, namely, y = x 3 / 2 and y = — x 3 / 2 . 
The equation sin(xy) — x 2 y 1 + e y = 0, on the other 
hand, cannot readily be solved for y in terms of x. See 
the end of §2.6 for more about implicit functions.) 
(a) Show that if F and y(x) are both assumed to be 
differentiable functions, then 

dy_ _ F x (x, y) 
dx F y (x, y) 

provided F y (x, y) # 0. 



(b) Use the result of part (a) to find dy /dx when y 
is defined implicitly in terms of x by the equa- 
tion x 3 — y 2 = 0. Check your result by explicitly 
solving for y and differentiating. 

35. Find dy /dx when y is defined implicitly by the equa- 
tion sin(xy) — x 2 y 7 + e y = 0. (See Exercise 34.) 

36. Suppose that you are given an equation of the form 

F(x,y,z) = Q, 

for example, something like x 3 z + y cosz + 
(sin y)/z = 0. Then we may consider z to be defined 
implicitly as a function z(x, y). 
(a) Use the chain rule to show that if F and z(x , y) are 
both assumed to be differentiable, then 



3z 
dx 



F x (x, y, z) 
F z (x,y,z)' 



dz 
dy 



F y (x,y,z) 
F z (x, y, z) 



(b) Use part (a) to find 3z/3x and 3z/3y where z is 
given by the equation xyz = 2. Check your result 
by explicitly solving for z and then calculating the 
partial derivatives. 

37. Find dz/dx and 3z/3y, where z is given implicitly by 
the equation 



, sin y 
x z + y cos z H = 0. 



(See Exercise 36.) 
38. Let 



f(x,y) 



2 

x y 
x 2 + y 2 

0 



if (x, y) # (0, 0) 
if(x,y) = (0,0) 



(a) Use the definition of the partial derivative to find 
f x (0, 0) and f y (0, 0). 

(b) Let a be a nonzero constant and let x(f) = 
(t, at). Show that / o x is differentiable, and find 
D(f o x)(0) directly. 

(c) Calculate Df(0, O)Z)x(O). How can you reconcile 
your answer with your answer in part (b) and the 
chain rule? 

Let w = f(x, y, z) be a differentiable function of x, y, and 
Z. For example, suppose that w = x + 2y + z. Regarding the 
variables x, y, and z as independent, we have dw/dx = 1 and 
dw/dy = 2. But now suppose that z = xy. Then x, y, and z 
are not all independent and, by substitution, we have that w = 
X + 2y + xy so that dw/dx = 1 + y and dw/dy = 2 + x. To 
overcome the apparent ambiguity in the notation for partial 
derivatives, it is customary to indicate the complete set of in- 
dependent variables by writing additional subscripts beside 



Chapter 2 | Differentiation in Several Variables 



the partial derivative. Thus, 



/9uA 



would signify the partial derivative of w with respect to x, 
while holding both y and z constant. Hence, x, y, and z are the 
complete set of independent variables in this case. On the other- 
hand, we would use (dw/dx) y to indicate thatx andy alone are 
the independent variables. In the case that w = x + 2y + z, 
this notation gives 



9 w 
dx 



( dw \ ( dw 
1, I I = 2, and I 

V dy J x , z V dz 



1. 



If Z = xy, then we also have 
dw\ 

= 1 + V, and 

dx /„ 



dw 
97 



2 + x. 



In this way, the ambiguity of notation can be avoided. Use this 
notation in Exercises 39-45. 



39. Let w = x + ly — lOz and z = x 2 + y 2 . 

dw\ / dw\ / dw\ f dw 

97j,,,; lay A*' \9z) x ,y' la* 

' dw 



(a) Find 



and 



dy 



(b) Relate (dw/dx) yz and (dw/dx) y by using the 
chain rule. 

40. Repeat Exercise 39 where w = x 3 + y 2 + z 3 and z = 

2x — 3y. 



41. Suppose s = x 2 y + xzw — z 2 and xyw — v 3 z + xz 
= 0. Find 



and 



ds 



42. Let U = F(P ', V, T) denote the internal energy of a 
gas. Suppose the gas obeys the ideal gas law P V = kT, 
where & is a constant. 
,'dU" 

(a) Find 



(b) Find 



(c) Find 



3T 

dU 
~df 

dU 
~d~P 



43 . Show that if x , y , z are related implicitly by an equation 
of the form F(x, y, z) = 0, then 



a*\ (dy\ (dz 

dy)z \dz) x \dxj v 



■1. 



This relation is used in thermodynamics. (Hint: Use 
Exercise 36.) 

44. The ideal gas law PV = kT, where £ is a constant, 
relates the pressure P, temperature T, and volume V 
of a gas. Verify the result of Exercise 43 for the ideal 
gas law equation. 

45. Verify the result of Exercise 43 for the ellipsoid 



ax 2 + by 2 + cz 2 = d 



where a, b, c, and d are constants. 



2.6 Directional Derivatives and the Gradient 

In this section, we will consider some of the key geometric properties of the 
gradient vector 

v/ = (i,i *L) 

\dx\ dx2 dx„J 

of a scalar-valued function of n variables. In what follows, n will usually be 2 
or 3. 

The Directional Derivative 

Let f(x , y) be a scalar- valued function of two variables. In §2.3, we understood the 
partial derivative j^{a,b) as the slope, at the point (a, b, f{a, b)), of the curve 
obtained as the intersection of the surface z = f(x, y) with the plane y = b. 
The other partial derivative ^-(a,b) has a similar geometric interpretation. How- 
ever, the surface z = f(x, y) contains infinitely many curves passing through 
(a, b, f(a, b)) whose slope we might choose to measure. The directional deriva- 
tive enables us to do this. 



2.6 | Directional Derivatives and the Gradient 



An alternative way to view ^(a,b) is as the rate of change of / as we move 
"infmitesimally" from a = (a, b) in the i-direction, as suggested by Figure 2.67. 
This is easy to see since, by the definition of the partial derivative, 

d f< « r f(a + h,b)-f(a,b) 
-(a, b) = hm 



dx h^O 



lim 



lim 



f((a,b) + (h,0))- f(a,b) 



f((a,b) + h(l,0))- f(a,b) 



o h 

r /(a + M)-/(a) 
= lim . 

h-+Q h 

Note that we are identifying the point (a, b) with the vector a = {a, b) = a\ + bj. 
Similarly, we have 

V, .. /(a + /ij)-/(a) 
— (a, b) = lim . 

dy h^o h 

Writing partial derivatives as we just have enables us to see that they are 
special cases of a more general type of derivative. Suppose v is any unit vector in 
R 2 . (The reason for taking a unit vector will be made clear later.) The quantity 

lim /(a + < ,v)-/(a) 

h^O h 

is nothing more than the rate of change of / as we move (infmitesimally) from a = 
(a , b) in the direction specified by v = (A, B) = Ai + By It's also the slope of the 
curve obtained as the intersection of the surface z = f{x,y) with the vertical plane 
B(x — a) — A(y — b) = 0. (See Figure 2.68.) We can use the limit expression in 
(1) to define the derivative of any scalar- valued function in a particular direction. 




Figure 2.67 Another way to view the 
partial derivative 9 // dx at a point. 



Figure 2.68 The directional derivative. 



Chapter 2 | Differentiation in Several Variables 



DEFINITION 6.1 Let X be open in R", /:XCR"->Ra scalar-valued 
function, and a e X . If v e R" is any unit vector, then the directional deriva- 
tive of / at a in the direction of v, denoted D v /(a), is 

/(a + /*v)-/(a) 
D v /(a) = hm 

h^O h 

(provided that this limit exists). 



EXAMPLE 1 Suppose f(x, y) = x 2 - 3xy + 2x - 5y. Then, if v = (v, w) e 
R 2 is any unit vector, it follows that 

f((0,0) + h(v,w))-f(0, 0) 



D y f(0, 0) = lim 



h^O h 

2 v 2 — 3h 2 vw + 2hv — 5hw 



= lim 

h^o h 

= lim(7ii; 2 — 3h vw + 2v — 5w) 
= 2v — 5w. 

Thus, the rate of change of / is 2v — 5w if we move from the origin in the 
direction given by v. The rate of change is zero if v = (5/V29, 2/V29) or 
(-5/V29, -2/V59). ♦ 

Consequently, we see that the partial derivatives of a function are just the "tip 
of the iceberg." However, it turns out that when / is differentiable, the partial 
derivatives actually determine the directional derivatives for all directions v. To 
see this rather remarkable result, we begin by defining a new function F of a 
single variable by 

F(0 = /(a-Mv). 

Then, by Definition 6. 1, we have 

n f( . Y /(a + rv)-/(a) F(t) - F(0) 
D v /(a) = hm = hm — = F (0). 

t^O t r^O t — 0 

That is, 

O v /(a) = ^/(a + *v)| f=0 . (2) 

The significance of equation (2) is that, when / is differentiable at a, we can apply 
the chain rule to the right-hand side. Indeed, let x(r) = a + fv. Then, by the chain 
rule, 

^/(a + fv) = Df(x)Dx(t) = D/(x)v. 
dt 

Evaluation at t = 0 gives 

D v /(a) = Z)/(a)v = V/(a).v. (3) 

The purpose of equation (3) is to emphasize the geometry of the situation. The 
result above says that the directional derivative is just the dot product of the 



2.6 | Directional Derivatives and the Gradient 



gradient and the direction vector v. Since the gradient is made up of the partial 
derivatives, we see that the more general notion of the directional derivative 
depends entirely on just the direction vector and the partial derivatives. To be 
more formal, we summarize this discussion with a theorem. 



THEOREM 6.2 Let X C R" be open and suppose /: X -> R is differentiable 
at a e X. Then the directional derivative D v /(a) exists for all directions (unit 
vectors) veR" and moreover, we have 

D v /(a) = V/(a).v. 



EXAMPLE 2 The function f(x, y) = x 2 — 3xy + 2x — 5y we considered in 
Example 1 has continuous partials and hence, by Theorem 3.5, is differentiable. 
Thus, Theorem 6.2 applies to tell us that, for any unit vector v = i>i + w\ 6 R 2 , 

D v /(0, 0) = V/(0, 0) • v = (f x (0, 0)i + /,(0, 0)j) • (vi + wj) 
= (2i-5j)-(ui+ W j) 

= 2v — 5w, 



as seen earlier. ♦ 

EXAMPLE 3 The converse of Theorem 6.2 does not hold. That is, a function 
may have directional derivatives in all directions at a point yet fail to be differ- 
entiable. To see how this can happen, consider the function /: R 2 -> R denned 
by 



/(*. y) = 



if (jc, y) * (0, 0) 

x L + y 4 

0 if(je,y) = (0,0) 



This function is not continuous at the origin. (Why?) So, by Theorem 3.6, it 
fails to be differentiable there; however, we claim that all directional derivatives 
exist at the origin. To see this, let the direction vector v be vi + wj. Hence, by 
Definition 6.1, we observe that 



D v /(0, 0) = lim 



f((0,0) + Kvi + wj))-f(0, 0) 
h 



= lim 

h^0 



hv(hw) 



{hvf + (hw) 4 
h 2 vw 2 



lim 

h^o h 2 (v 2 + h 2 w 4 ) 

,2 „,„2 



= lim 



vw 



h^0 V 2 + h 2 w 4 



vw 



w 

V 



Chapter 2 | Differentiation in Several Variables 



Thus, the directional derivative exists whenever v 7^ 0. When v = 0 (in which 
case v = j), we, again, must calculate 

/((0, 0) + fcj)-/(0, 0) 



Dj/(0, 0)=lim 



h-*o h 

= hm 

fc->-0 

0-0 
= hm = 0. 

ft->-o h 

Consequently, this directional derivative (which is, in fact, df/dy) exists as well. 

♦ 

The reason we have restricted the direction vector v to be of unit length in our 
discussion of directional derivatives has to do with the meaning of D v /(a), not 
with any technicalities pertaining to Definition 6.1 or Theorem 6.2. Indeed, we 
can certainly define the limit in Definition 6. 1 for any vector v, not just one of unit 
length. So, suppose w is an arbitrary nonzero vector in R" and / is differentiable. 
Then the proof of Theorem 6.2 goes through without change to give 

hm = V / (a) • w. 

h-*o h 

The problem is as follows: If w = kv for some (nonzero) scalar k, then 

hm = V / (a) • w 

h-f0 h 

= V/(a)-(*v) 
= *(V/(a)-v) 

n /(a + Av)-/(a) 



That is, the "generalized directional derivative" in the direction of kv is k times 
the derivative in the direction of v. But v and kv are parallel vectors, and it is 
undesirable to have this sort of ambiguity of terminology. So we avoid the trouble 
by insisting upon using unit vectors only (i.e., by allowing k to be ±1 only) when 
working with directional derivatives. 



Gradients and Steepest Ascent 

Suppose you are traveling in space near the planet Nilrebo and that one of your 
spaceship's instruments measures the external atmospheric pressure on your ship 
as a function f(x, y, z) of position. Assume, quite reasonably, that this function 
is differentiable. Then Theorem 6.2 applies and tells us that if you travel from 
point a = (a , b, c) in the direction of the (unit) vector u = ui + uj + 10k, the rate 
of change of pressure is given by 

D u /(a) = V/(a).u. 

Now, we ask the following: In what direction is the pressure increasing the most? 
If 9 is the angle between u and the gradient vector V/(a), then we have, by 
Theorem 3.3 of §1.3, that 



D n /(a) = ||V/(a)|| ||u|| costf = ||V/(a)|| cos#, 



2.6 | Directional Derivatives and the Gradient 163 

since u is a unit vector. Because — 1 < cos 6 < 1 , we have 

-||V/(a)|| < D u /(a)< ||V/(a)||. 

Moreover, cos 9 = 1 when 9 = 0 and cos 9 = — 1 when 9 = it . Thus, we have 
established the following: 

THEOREM 6.3 The directional derivative D„f(a) is maximized, with respect to 
direction, when u points in the same direction as V/(a) and is minimized when u 
points in the opposite direction. Furthermore, the maximum and minimum values 
of D u /(a) are || V/(a)|| and -|| V/(a)||, respectively. 

EXAMPLE 4 If the pressure function on Nilrebo is 

f(x, y, z) = 5x 2 + 7y 4 + x 2 z 2 arm, 

where the origin is located at the center of Nilrebo and distance units are measured 
in thousands of kilometers, then the rate of change of pressure at (1,-1,2) 
in the direction of i + j + k may be calculated as V/(l, — 1, 2) • u, where u = 
(i + j + k)/\/3. (Note that we normalized the vector i + j + k to obtain a unit 
vector.) Using Theorem 6.2, we compute 

A./(l,-l,2) = V/(l,-l,2).u 

i + j + k 

= (18i - 28j + 4k) V— 

18-28 + 4 



V3 



2\/3 atm/Mm. 



Additionally, in view of Theorem 6.3, the pressure will increase most rapidly 
in the direction of V/(l , —1,2), that is, in the 

18i-28j+4k 9i-14j + 2k 



||18i-28j+4k|| V28T 

direction. Moreover, the rate of this increase is 

|| V/(l, -1,2)|| = 2^/281 atm/Mm. ♦ 

Theorem 6.3 is stated in a manner that is independent of dimension — that is, so 
that it applies to functions /: X C R" —> R for any n > 2. In the case n = 2, there 
is another geometric interpretation of Theorem 6.3: Suppose you are mountain 
climbing on the surface z = fix, y). Think of the value of / as the height of the 
mountain above (or below) sea level. If you are equipped with a map and compass 
(which supply information in the xy-plane only), then if you are at the point on 
the mountain with xy-coordinates (map coordinates) (a, b), Theorem 6.3 says 
that you should move in the direction parallel to the gradient V/(a , b) in order to 
climb the mountain most rapidly. (See Figure 2.69.) Similarly, you should move 
in the direction parallel to — V f(a , b) in order to descend most rapidly. Moreover, 
the slope of your ascent or descent in these cases is || V/(a, b)\\ . Be sure that you 
understand that V/(a, b) is a vector in R 2 that gives the optimal north-south, 
east-west direction of travel. 



1 64 Chapter 2 | Differentiation in Several Variables 





Tangent Planes Revisited 

In §2.1, we indicated that not all surfaces can be described by equations of the 
form z = f(x, y). Indeed, a surface as simple and familiar as the sphere is not the 
graph of any single function of two variables. Yet the sphere is certainly smooth 
enough for us to see intuitively that it must have a tangent plane at every point. 
(See Figure 2.70.) 

How can we find the equation of the tangent plane? In the case of the unit 
sphere x 2 + y 2 + z 2 = 1, we could proceed as follows: First decide whether the 
point of tangency is in the top or bottom hemisphere. Then apply equation (4) of 



Figure 2.70 A sphere and one of §2.3 to the graph of z = — x 2 — y 2 or z = —y/\ — x 2 — y 2 , as appropriate. 

its tangent planes. The calculus is tedious but not conceptually difficult. However, the tangent planes 

to points on the equator are all vertical and so equation (4) of §2.3 does not apply. 
(It is possible to modify this approach to accommodate such points, but we will 
not do so.) In general, given a surface described by an equation of the form 
F(x, v, z) = c (where c is a constant), it may be entirely impractical to solve 
for z even as several functions of x and y. Try solving for z in the equation 
xyz + ye" — x 2 + yz 2 = 0 and you'll see what we mean. We need some other 
way to get our hands on tangent planes to surfaces described as level sets of 
functions of three variables. 

To get started on our quest, we present the following result, interesting in its 
own right: 



THEOREM 6.4 Let X C R" be open and /: X -> R be a function of class C 1 . 
If x 0 is a point on the level set S = {x e X \ f(x) = c}, then the vector V/(x 0 ) 
is perpendicular to S. 



2.6 | Directional Derivatives and the Gradient 



PROOF We need to establish the following: If v is any vector tangent to S at xo, 
then V/(xo) is perpendicular to v (i.e., V/(xo) • v = 0). By a tangent vector to S 
at xo, we mean that v is the velocity vector of a curve C that lies in S and passes 
through Xq. The situation in R 3 is pictured in Figure 2.71. 




Figure 2.71 The level set surface 
S = {x | /(x) = c}. 



Thus, let C be given parametrically by x(t) = (xi(/), X2(t), . . . , x„(t)), where 
a < t < b and x(fo) = xo for some number to in (a, b). (Then, if v is the velocity 
vector at x 0 , we must have x'(f 0 ) = v. See §3.1 for more about velocity vectors.) 
Since C is contained in S, we have 

/(x(0) = /(*i(0. *2(0. • ■ • . *»(0) = c - 

Hence, 

j t imm = jfc\ = o. (4) 

On the other hand the chain rule applied to the composite function fox: 
(a,b) -> R tells us 

^[/(x(f))] = V/(x(0)-x'(0- 
Evaluation at to and equation (4) let us conclude that 

V/(x(/ 0 ))-x'(*o) = V/(x 0 )-v = 0, 
as desired. ■ 



Here's how we can use the result of Theorem 6.4 to find the plane tan- 
gent to the sphere x 2 + y 2 + z 2 = 1 at the point ^— 0, ^J. From §1.5, we 

know that a plane is determined uniquely from two pieces of information: (i) a 
point in the plane and (ii) a vector perpendicular to the plane. We are given a 

point in the plane in the form of the point of tangency ^— -7^, 0, ^=V As for 

a vector normal to the plane, Theorem 6.4 tells us that the gradient of the func- 
tion f(x, y, z) = x 2 + y 2 + z 2 that defines the sphere as a level set will do. We 
have 

V f(x, y, z) = 2xi + 2y\ + 2zk, 

so that 

v/ (-7!'°'7!) = ^' +V5k - 



Chapter 2 | Differentiation in Several Variables 



Hence, the equation of the tangent plane is 

V2f, + -L) + ^--L)=0, 



= 0, 



or 



X = y/2. 



In general, if S is a surface in R 3 defined by an equation of the form 

f(x,y,z) = c, 

then if xo £ X, the gradient vector V/(xo) is perpendicular to S and, conse- 
quently, if nonzero, is a vector normal to the plane tangent to S at x 0 . Thus, 
the equation 



V/(x 0 )-(x-x 0 ) = 0 

or, equivalently, 

f x (x 0 , y 0 , z 0 )(x - x 0 ) + f y (x 0 , y 0 , zo)(y - v 0 ) 

+ fz(*o, yo, zo)(z - zo) = 0 
is an equation for the tangent plane to S at x 0 . 



(5) 



(6) 



Note that formula (5) can be used in R" as well as in R 3 , in which case it 
defines the tangent hyperplane to the hypersurface S C R" defined by f{x\ , 
X2, ■ ■ ■ , x n ) = c at the point xo e S. 

EXAMPLE 5 Considerthe surface S defined by the equation* 3 y — yz 2 + z 5 = 
9. We calculate the plane tangent to S at the point (3,-1,2). 
To do this, we define f(x, y, z) = x 3 y — yz 2 + z 5 - Then 

V/(3, -1,2)= (3x 2 yi + (x 3 - z 2 )\ + (5 Z 4 - 2yz)k)\ (3 _ h2) 

= -21\ + 23j + 84k 

is normal to S at (3, —1, 2) by Theorem 6.4. Using formula (6), we see that the 
tangent plane has equation 



or, equivalently, 



-270 - 3) + 23(y + 1) + 84(z - 2) = 0 



-27x + 23y + 84z = 64. 



EXAMPLE 6 Consider the surface defined by z 4 = x 2 + y 2 . This surface is 
the level set (at height 0) of the function 



f{x, y, Z ) = x 2 + y 2 - z 4 . 



The gradient of / is 



Vf(x, y,z) = 2xi + 2yj- 4z 3 k. 



2.6 | Directional Derivatives and the Gradient 




Figure 2.72 The 

surface of Example 6. 



Note that the point (0, 0, 0) lies on the surface. However, V/(0, 0, 0) = 0, which 
makes the gradient vector unusable as a normal vector to a tangent plane. Thus, 
formula (6) doesn't apply. What we conclude from this example is that the surface 
fails to have a tangent plane at the origin, a fact that is easy to believe from the 
graph. (See Figure 2.72.) ♦ 

EXAMPLE 7 The equation x 2 + y 2 + z 2 + w 2 = 4 defines a hypersphere of 

radius 2 in R 4 . We use formula (5) to determine the hyperplane tangent to the 
hypersphere at (— 1 , 1 , 1,-1). 

The hypersphere may be considered to be the level set at height 4 of the 
function f(x, y, z, w) = x 2 + y 2 + z 2 + w 2 , so that the gradient vector is 

V f(x, y, z, w) = (2x, 2y, 2z, 2w), 

so that 

V/(-l, 1,1,-1) = (-2, 2, 2, -2). 
Using formula (5), we obtain an equation for the tangent hyperplane as 
(-2,2,2, -2)-(x + l,y- 1, z - 1, w + 1) = 0 

or 

-2(x + 1) + 2(y - 1) + 2(z - 1) - 2(w + 1) = 0. 
Equivalently, we have the equation 

x - y - z + w + 4 = 0. ♦ 



EXAMPLE 8 We determine the plane tangent to the paraboloid z 



- v2 



3y 



at the point (—2, 1, 7) in two ways: (i) by using formula (4) in §2.3, and (ii) by 
using our new formula (6). 

First, the equation z = x 2 + 3y 2 explicitly describes the paraboloid as the 
graph of the function f(x, y) = x 2 + 3y 2 , that is, by an equation of the form 
z = fix, y). Therefore, formula (4) of §2.3 applies to tell us that the tangent 
plane at (—2, 1, 7) has equation 

z = f(-2, 1) + f x (-2, l)(x + 2) + f y (-2, l)(y - 1) 

or, equivalently, 

z = 7 - 4(jc + 2) + 6(y - 1). (7) 

Second, if we write the equation of the paraboloid as x 2 + 3y 2 — z = 0, 
then we see that it describes the paraboloid as the level set of height 0 of the 
three-variable function F(x, y, z) = x 2 + 3y 2 — z. Hence, formula (6) applies 
and indicates that an equation for the tangent plane at (—2, 1, 7) is 

F x (-2, 1, 7)(jc + 2) + F y {-2, 1, 7)(y - 1) + F z (-2, 1, 7)(z - 7) = 0 



or 



-4(x + 2) + 6(y- 1) - l(z - 7) 
As can be seen, equation (7) agrees with equation (8). 



(8) 

♦ 



Example 8 may be viewed in a more general context. If S is the surface in R 3 
given by the equation z = f(x, y) (where / is differentiate), then formula (4) of 
§ 2 . 3 tells us that an equation for the plane tangent to S at the point (a , b , / {a , b)) is 



z = f(a, b) + f x (a, b)(x -a)+ Ua, b)(y - b). 



Chapter 2 | Differentiation in Several Variables 



(-2,2,6) 




Figure 2.73 The two-sheeted 
hyperboloid z 2 /4 — x 2 — y 2 = 1. 
The point (—2, 2, 6) lies on the 
sheet given by z = 2^/x 2 + y 2 + 1, 
and the point (1,1, — 2^/3) lies on 
the sheet given by 
z = -2jx 2 + y 2 +l. 



1. 



At the same time, the equation for S may be written as 

f(x, y) - z = 0. 

Then, if we let F(x, y, z) = f(x, y) — z, we see that S is the level set of F at 
height 0. Hence, formula (6) tells us that the tangent plane at (a, b, f(a, b)) is 

F x (a, b, f(a, b))(x - a) + F y (a, b, f(a, b))(y - b) 

+ F e (a, b, f(a, b))(z - f(a, b)) = 0. 

By construction of F, 

dF _df dF _ df dF 

dx dx ' dy dy ' dz 

Thus, the tangent plane formula becomes 

f x (a, b)(x -a) + f y (a, b)(y - b) - (z - f(a, b)) = 0. 

The last equation for the tangent plane is the same as the one given above by 
equation (4) of §2.3. 

The result shows that equations (5) and (6) extend the formula (4) of §2.3 to 
the more general setting of level sets. 

The Implicit Function and Inverse Function 

Theorems (optional) 

We have previously noted that not all surfaces that are described by equations of 
the form F(x ,y,z) = c can be described by an equation of the form z = f(x , y). 
We close this section with a brief — but theoretically important — digression about 
when and how the level set {(x, y, z) | F(x, y, z) = c} can also be described as 
the graph of a function of two variables, that is, as the graph of z = f(x, y). 
We also consider the more general question of when we can solve a system of 
equations for some of the variables in terms of the others. 
We begin with an example. 



EXAMPLE 9 Consider the hyperboloid z 2 /4 - x 2 
described as the level set (at height 1) of the function 



y = 1, which may be 



F{x, y, z) = 



y 



(See Figure 2.73.) This surface cannot be described as the graph of an equation 
of the form z = f(x, y), since particular values for x and y give rise to two values 
for z. Indeed, when we solve for z in terms of x and y, we find that there are two 
functional solutions: 



x 2 + y 2 + 1 and z 



= -2/ 



x 2 + y 2 + 1. 



(9) 



On the other hand these two solutions show that, given any particular point 
(*o, yo, zo) of the hyperboloid, we may solve locally for z in terms of x and y. 
That is, we may identify on which sheet of the hyperboloid the point {xq, yo, Zo) 
lies and then use the appropriate expression in (9) to describe that sheet. ♦ 

Example 9 prompts us to pose the following question: Given a surface S, 
described as the level set {(x, y, z) | F(x, y, z) = c], can we always determine at 
least a portion of S as the graph of a function z = f(x , y)? The result that follows, 



2.6 | Directional Derivatives and the Gradient 



a special case of what is known as the implicit function theorem, provides 
relatively mild hypotheses under which we can. 



THEOREM 6.5 (The implicit function theorem) Let F:XcR"->R be 
of class C 1 and let a be a point of the level set S = {x 6 R" | F(x) = c}. If 
F Xn (a) 0, then there is a neighborhood U of (a\, ci2, . . . , a„_i) in R n_1 , a 
neighborhood V of a n in R, and a function f:U<^ R" _1 -> V of class C 1 
such that if (xi, X2, ■ ■ ■ , x n -\) e U andx„ e V satisfy F(x\, X2, ■ • ■ , x n ) = c (i.e., 
(x\,x 2 , x„) e S), thenx„ = f(xi,x 2 , x„_i). 



The significance of Theorem 6.5 is that it tells us that near a point a € S 
such that dF/dx n ^ 0, the level set S given by the equation F{x\, . . . , x n ) = c 
is locally also the graph of a function x„ = /(xj , . . . , x„_i). In other words, we 
may solve locally for x n in terms of x\, . . . , x„_i, so that S is, at least locally, a 
differentiable hypersurface in R" . 

EXAMPLE 10 Returning to Example 9, we recall that the hyperboloid is the 
level set (at height 1) of the function F(x, y, z) = z 2 /4 — x 2 — y 2 . We have 



dF 

9z" 



Note that for any point (xo, yo> Zo) in the hyperboloid, we have |zol > 2. Hence, 
9F.(xo, yo, zo) 0. Thus, Theorem 6.5 implies that we may describe a portion 
of the hyperboloid near any point as the graph of a function of two variables. This 
is consistent with what we observed in Example 9. ♦ 

Of course, there is nothing special about solving for the particular vari- 
able x„ in terms of Suppose a is a point on the level set S de- 
termined by the equation F(x) = c and suppose VF(a) / 0. Then F Xj (a) 0 for 
some i . Hence, we can solve locally near a for x,- as a differentiable function of 
x\, . . . , x,_i, Xi+i, . . . , x„. Therefore, S is locally a differentiable hypersurface 
inR". 



EXAMPLE 11 Let S denote the ellipsoid x 2 /4 + y 2 /36 + z 2 /9 
is the level set (at height 1) of the function 



1. Then S 



n*,y,z)=^ + y - + 



At the point (V2, V6, V3), we have 



dF 

~dz 



2z 



(V2.V6.V3) 



2^3 



(V2,V6,V3) 



Thus, S may be realized near (V2, V6, V3) as the graph of an equation of the 

form z = f(x, y), namely, z = 3^/1 - x 2 /4 — y 2 /36. At the point (0, —6, 0), 
however, we see that dF /dz vanishes. On the other hand, 



dF 

97 



(0,-6,0) 



2 ± 

36 



(0,-6,0) 



^0. 



Consequently, near (0, —6, 0), the ellipsoid may be described by solving for y as 
a function of x and z, namely, y = —6^/ 1 — x 2 /4 — z 2 /9. ♦ 



Chapter 2 | Differentiation in Several Variables 



EXAMPLE 1 2 Consider the set of points S defined by the equation x 2 z 2 — y = 
0. Then S is the level set at height 0 of the function F(x, y, z) = x 2 z 2 — y. Note 
that 

VF(x,y,z) = (2xz 2 , -l,2x 2 z). 

Since dF /dy never vanishes, we see that we can always solve for y as a function 
of x and z. (This is, of course, obvious from the equation.) On the other hand, near 
points where x and z are nonzero, both dF/dx and dF/dz are nonzero. Hence, 
we can solve for either x or z in this case. For example, near (1 , 1 , — 1), we have 

x= H and z= -Ii- 

As just mentioned, Theorem 6.5 is actually a special case of a more general 
result. In Theorem 6.5 we are attempting to solve the equation 

F(xi, x 2 , . . . , x n ) = c 

for x n in terms of x\, . . . , x n -\. In the general case, we have a system of m 
equations 

F\(x\, . . . , x„, yi, . . . , y m ) = c 1 
F 2 (xi,...,x„,yi,...,y m ) = c 2 

^mv^l ' • ■ ■ ' 3^1) ■ ■ ■ ' y?n) 

and we desire to solve the system for yi , . . . , y m in terms of x\ , . . . , x„. Using vec- 
tor notation, we can also write this system as F(x, y) = c, where x = (xi, . . . , x n ), 
y = (yi, . . . , y m ), c = (ci , . . . , c m ), and Fi, ... , F m make up the component 
functions of F. With this notation, the general result is the following: 



THEOREM 6.6 (The implicit function theorem, general case) Suppose 
F: A ->• R m is of class C 1 , where A is open in R" +m . Let (a, b) = (a\, . . . , a„, 
b\, . . . , b m ) £ A satisfy F(a, b) = c. If the determinant 



r 3Fi 

dy 



A(a, b) = det 



(a,b) 



9F, 

dy, 



-(a, b) 



dF m 



L dy 



-(a, b) 



dF m 

dy r . 



(a, b) 



^0, 



then there is a neighborhood U of a in R" and a unique function f : U —> R m of 
class C 1 such that f(a) = b and F(x, f(x)) = c for all x e U. In other words, we 
can solve locally for y as a function f(x). 



EXAMPLE 13 We show that, near the point (x\, x%, x^, y\, y 2 ) = (— 1, 1, 1, 
2, 1), we can solve the system 

x\y 2 + x 2 y\ = 1 
xfx 3 yi + x 2 y{ = 3 
for y\ and y 2 in terms of x\ , x 2 , x%. 



2.6 | Directional Derivatives and the Gradient 



We apply the general implicit function theorem (Theorem 6.6) to the system 

IFi(xi, x 2 , x 3 , y\,yi) = x\y 2 + x 2 yi = 1 
F 2 (xi,x 2 , x 3 , y\,y 2 ) = xfx 3 yi + x 2 y\ = 3 
The relevant determinant is 



A(-l, 1, 1,2, 1) = det 



= det 



= det 



3Fi dFi 

dyi dy 2 

dF 2 dF 2 

dyi dy 2 

x 2 X\ 



(X1,.T 2 ,X3,3'1,}' 2 )=(- 1,1, 1,2,1) 



xfxj, 3x 2 yj 



(xi ,a- 2 , x 3 , vi , y 2 )={- 1 , 1 , 1 ,2, 1) 

4/0. 



Hence, we may solve locally, at least in principle. 

We can also use the equations in (11) to determine, for example, 

dy 2 

(— 1, 1, 1), where we treat x\, x 2 , x 3 as independent variables and y\ and 

dx\ 

y 2 as functions of them. 

Differentiating the equations in (1 1) implicitly with respect to X\ and using 
the chain rule, we obtain 



yi + x x - — Vx 2 — 

ax\ ax\ 



0 



Ixix^yi +x\xt,- r-3x 2 y; 



dy 2 



= 0 



.2„ 

Now, let (xi , x 2 , Xi, y\ , y 2 ) — (— 1, 1 , 1,2, 1), so that the system becomes 



f*(-l,l,l)-|^(-l,l,l) = -l 
3xi 6x\ 

^l(-l,l,l) + 3^(-l,l,l) = 4 
3xi dxi 

3y2 5 

We may easily solve this last system to find that — (—1,1,1) = —. ♦ 

3xi 4 

Now, suppose we have a system of n equations that defines the variables 
yi , . . . , y n in terms of the variables xi, . . . , x„, that is, 



yi = /i(xi, . . . ,x„) 
yi = fi(xu ■ . . ,x n ) 

y„ = fn(X\,...,X n ) 



(12) 



Note that the system given in (12) can be written in vector form as y = f(x). The 
question we ask is, when can we invert this system? In other words, when can we 



Chapter 2 | Differentiation in Several Variables 



solve for x\ , . . . ,x n in terms of yi , 
function g so that x = g(y)? 

The solution is to apply Theorem 6.6 to the system 



y n , or, equivalently, when can we find a 



F 2 (xi, 



,x n ,y\, 
x„,y { , 

.x n ,y\, 



y„) = 0 

y„) = o 

Jn) = 0 



where Ff(jti, . . . , x n , y\, . . . , y„) = fi(xi, . . . , x n ) — y,-. (In vector form, we are 
setting F(x, y) = f(x) — y.) Then solvability for x in terms of y near x = a, y = b 
is governed by the nonvanishing of the determinant 



detDf(a) = det 



This determinant is also denoted by 

3(/i- 



3xi 



dxt 



(a) 



(a) 



dx n 

dfn 
dx n 



(a) 



(a) 



d(x u . . . ,x„) 

and is called the Jacobian of f = (/i, . . . , /„). A more precise and complete 
statement of what we are observing is the following: 



THEOREM 6.7 (The inverse function theorem) Suppose f = (f\ /„) 

is of class C 1 on an open set A c R" . If 

then there is an open set U c R" containing a such that f is one-one on U, the 
set V = f(U) is also open, and there is a uniquely determined inverse function 
g: V -> U to f, which is also of class C 1 . In other words, the system of equations 
y = f(x) may be solved uniquely as x = g(y) for x near a and y near b. 



det Df(a) 



3(/i,..../») 



3(xi, . . . , x n ) 



EXAMPLE 14 Consider the equations that relate polar and Cartesian coordi- 
nates: 

' x = r cos# 
y = r sin 6 

These equations define x and y as functions of r and 6. We use Theorem 6.7 to 
see near which points of the plane we can invert these equations, that is, solve for 
r and 9 in terms of x and y. 

To use Theorem 6.7, we compute the Jacobian 



d(x, y) 
d(r, 9) 



cos 6 —rsmO 
sin 0 r cos 9 



r. 



Thus, we see that, away from the origin (r = 0), we can solve (locally) for r and 9 
uniquely in terms of x and y. At the origin, however, the inverse function theorem 



2.6 I Exercises 173 



does not apply. Geometrically, this makes perfect sense, since at the origin the 
polar angle 9 can have any value. ♦ 



2.6 Exercises 



1 . Suppose f(x , y , z) is a differentiable function of three 
variables. 

(a) Explain what the quantity V/(x, y, z) • (— k) rep- 
resents. 

(b) How does V/(x, y, z) • (-k) relate to 3//3z? 

In Exercises 2—8, calculate the directional derivative of the 
given function f at the point a in the direction parallel to the 
vector u. 



2. f(x, y) = e v sinx, a = o), u 



31- j 



10 



3. f(x, y) = x 2 - 2x 3 y + 2y\ a = (2, -1), u = 



4. f{x, y) = 



1 



-,a = (3, -2),u = i-j 



(x 2 + y 2 )' 

5. /(x, y) = e x - x 2 y, a = (1, 2), u = 2i + j 



6. f(x, y, z) = xyz, a = (-1, 0, 2), u 



2k- i 

~7f 



7. /(x, y, z) = e~ 
8- f(x,y,z) 



3k 

9. For the function 

/(*. y) 



3z 2 + l 



+z \a = (l,2,3),u = i + j + k 
, a = (2, -1,0), u = i-2j + 



x\y\ 



0 



if (x, y) ? (0, 0) 
if(x,y) = (0, 0) 



(a) calculate f x (0, 0) and f y (0, 0). (You will need to 
use the definition of the partial derivative.) 

(b) use Definition 6.1 to determine for which unit 
vectors v = ui + urj the directional derivative 
D v /(0, 0) exists. 

(c) use a computer to graph the surface z = f(x, y). 
1 0. For the function 



XV 



Jx 2 + . 



if(x,y)/(0, 0) 
if(x,y) = (0, 0) 



(a) calculate / v (0, 0) and /,(0, 0). 



(b) use Definition 6.1 to determine for which unit 
vectors v = vi + w\ the directional derivative 
D v /(0, 0) exists. 

(c) use a computer to graph the surface z = f(x, y). 

1 1 . The surface of Lake Erehwon can be represented by a 
region D in the jcv-plane such that the lake's depth (in 
meters) at the point (x, y) is given by the expression 
400 — 3x 2 y 2 . If your calculus instructor is in the wa- 
ter at the point (1, —2), in which direction should she 
swim 

(a) so that the depth increases most rapidly (i.e., so 
that she is most likely to drown)? 

(b) so that the depth remains constant? 

12. A ladybug (who is very sensitive to temperature) is 
crawling on graph paper. She is at the point (3,7) and 
notices that if she moves in the i-direction, the tem- 
perature increases at a rate of 3 deg/cm. If she moves 
in the j -direction, she finds that her temperature de- 
creases at a rate of 2 deg/cm. In what direction should 
the ladybug move if 

(a) she wants to warm up most rapidly? 

(b) she wants to cool off most rapidly? 

(c) she desires her temperature not to change? 

13. You are atop Mt. Gradient, 5000 ft above sea level, 
equipped with the topographic map shown in Fig- 
ure 2.74. A storm suddenly begins to blow, necessitat- 
ing your immediate return home. If you begin heading 
due east from the top of the mountain, sketch the path 
that will take you down to sea level most rapidly. 

14. It is raining and rainwater is running off an ellipsoidal 
dome with equation Ax 2 + y 2 + 4z 2 = 16, where 
z > 0. Given that gravity will cause the raindrops to 
slide down the dome as rapidly as possible, describe 
the curves whose paths the raindrops must follow. 
(Hint: You will need to solve a simple differential 
equation.) 

15. Igor, the inchworm, is crawling along graph paper in 
a magnetic field. The intensity of the field at the point 
(x, y) is given by M(x, y) = 3x 2 + y 2 + 5000. If Igor 
is at the point (8, 6), describe the curve along which he 
should travel if he wishes to reduce the field intensity 
as rapidly as possible. 

In Exercises 16— 19, find an equation for the tangent plane to 
the surface given by the equation at the indicated point (xo , yo , 
zo)- 



1 74 Chapter 2 | Differentiation in Several Variables 




Figure 2.74 The topographic map of Mt. Gradient in Exercise 13. 



16. x 1 + y 3 + ; 3 = 7, (xq, y 0 , zo) = (0, -1, 2) 

17. ze y cosx = 1, (x 0 , yo, zo) = (x, 0, -1) 

18. 2xz + yz- x 2 y + 10 = 0, Oo, yo, zo) = (1, -5, 5) 

19. 2xy 2 = 2z 2 - xyz, (x 0 , y 0 , Zo) = (2, -3, 3) 

20. Calculate the plane tangent to the surface whose equa- 
tion is x 2 — 2y 2 + 5xz = 7 at the point (—1, 0, — |)in 
two ways: 

(a) by solving for z in terms of x and y and using 
formula (4) in §2.3 

(b) by using formula (6) in this section. 

21. Calculate the plane tangent to the surface xsiny + 
xz 2 = 2e yz at the point (2, j, 0) in two ways: 

(a) by solving for x in terms of y and z and using a 
variant of formula (4) in §2.3 

(b) by using formula (6) in this section. 

22. Find the point on the surface x 3 — 2y 2 + z 2 = 21 
where the tangent plane is perpendicular to the line 
given parametrically as x = 3f — 5, v = 2t + 7, z = 
1 - V2f. 

23. Find the points on the hyperboloid 9x 2 — 45y 2 + 
5z 2 = 45 where the tangent plane is parallel to the 
plane x + 5y — 2z = 7. 

24. Show that the surfaces z = lx 2 — \2x — 5y 2 and 



xyz 



2 intersect orthogonally at the point (2, 1 , — 1). 



25. Suppose that two surfaces are given by the equations 



Moreover, suppose that these surfaces intersect at the 
point (jco, yo, Zo)- Show that the surfaces are tangent at 
(x 0 , yo, Zo) if and only if 

VF(x 0 , yo, zo) x VG(i 0 , y 0 , zo) = 0. 

26. Let S denote the cone x 2 + Ay 2 = z 2 ■ 

(a) Find an equation for the plane tangent to S at the 
point (3, —2, —5). 

(b) What happens if you try to find an equation for a 
tangent plane to S at the origin? Discuss how your 
findings relate to the appearance of S. 

27. Consider the surface S defined by the equation x 3 — 

x 2 y 2 + Z 2 = 0. 

(a) Find an equation for the plane tangent to S at the 
point (2, -3/2, 1). 

(b) Does S have a tangent plane at the origin? Why or 
why not? 

If a curve is given by an equation of the form fix, y) = 0, then 
the tangent line to the curve at a given point (xo, yo) ° n it niay 
be found in two ways: (a) by using the technique of implicit 
differentiation from single-variable calculus and (b) by using 
a formula analogous to formula (6). In Exercises 28-30, use 
both of these methods to find the lines tangent to the given 
curves at the indicated points. 

28. x 2 + y 2 = 4, (xo, y 0 ) = (-V2, V2) 

29. y 3 =x 2 +x\(x 0 ,yo) = (1,^/2) 



F(x, y, z) = c and G(x, y, z) = k. 



30. x 5 + 2xy + y 3 = 16, (x 0 , y 0 ) = (2, -2) 



2.6 I Exercises 175 



Let C be a curve in R 2 given by an equation of the form 
f(x, y) = 0. The normal line to C at a point (xo, yo) on it 
is the line that passes through (xo, yo) and is perpendicular 
to C (meaning that it is perpendicular to the tangent line to 
C at (xo, yo)). In Exercises 31-33, find the normal lines to 
the given curves at the indicated points. Give both a set of 
parametric equations for the lines and an equation in the form 
Ax + By = C. (Hint: Use gradients.) 

31. x 2 -y 2 = 9,(x 0 ,y 0 ) = (5, -4) 

32. x 2 - x 3 = y 2 , (x 0 , yo) = (-1, 72) 

33. x 3 - 2xy + y 5 = 11, (x 0 , y 0 ) = (2, -1) 

34. This problem concerns the surface defined by the equa- 
tion 

x 3 z + x 2 y 2 + sin(yz) = -3. 

(a) Find an equation for the plane tangent to this sur- 
face at the point (—1, 0, 3). 

(b) The normal line to a surface S in R 3 at a point 
(xo, yo, Zo) on it is the line that passes through 
(xo, yo, Zo) and is perpendicular to S. Find a set 
of parametric equations for the line normal to the 
surface given above at the point (—1,0, 3). 

35. Give a set of parametric equations for the normal line to 
the surface defined by the equation e xy + e xz — 2e yz = 
0 at the point (-1,-1,-1). (See Exercise 34.) 

36. Give a general formula for parametric equations for 
the normal line to a surface given by the equation 
F(x, y, z) = 0 at the point (xo, yo, zo) on the surface. 
(See Exercise 34.) 

37. Generalizing upon the techniques of this section, 
find an equation for the hyperplane tangent to 
the hypersurface sinxi + cosx2 + sinx3 + cos X4 + 
sinxs = — 1 at the point (jr, jt, 3jt/2, 2jt, 2jt) e R 5 . 

38. Find an equation for the hyperplane tangent to the 
(« — l)-dimensional ellipsoid 

2, 0 2,0 2 , , 2 »(" + 1) 

X[ + 2x 2 + 3x 3 H h nx n = 

at the point (-1, -1,..., -1) e R". 

39. Find an equation for the tangent hyperplane to the (n — 
l)-dimensional sphere x\ + x\ + ■ ■ ■ + x 2 = 1 in R" 
at the point (1 j+Jn, \/-Jn, . . . , l/Vn, — l/*/n). 

Exercises 40—49 concern the implicit function theorems and 
the inverse function theorem (Theorems 6.5, 6.6, and 6. 7). 

40. Let S be described by z 2 y 3 + x 2 y = 2. 

(a) Use the implicit function theorem to determine 
near which points S can be described locally as 
the graph of a C 1 function z = /(x, y). 



(b) Near which points can S be described (locally) as 
the graph of a function x = g(y, z)? 

(c) Near which points can S be described (locally) as 
the graph of a function y = h(x, z)? 

41 . Let S be the set of points described by the equation 

sinxy + e xz + x 3 y = 1. 

(a) Near which points can we describe S as the graph 
of a C 1 function z = f(x, y)? What is f{x, y) in 
this case? 

(b) Describe the set of "bad" points of S, that is, the 
points (xo, yo, zo) € S where we cannot describe 
S as the graph of a function z = f(x, y). 

(c) Use a computer to help give a complete picture of 
S. 

42. Let F(x, y) = c define a curve C in R 2 . Suppose 
(xo, yo) is a point ofC such that VF(xo, yo) / 0. Show 
that the curve can be represented near (xo, yo) as either 
the graph of a function y = f(x) or the graph of a 
function x = g(y). 

43. Let F(x, y) = x 2 — y 3 , and consider the curve C de- 
fined by the equation F(x, y) = 0. 

(a) Show that (0, 0) lies on C and that F y (0, 0) = 0. 

(b) Can we describe C as the graph of a function 
y = /(x)? Graph C. 

(c) Comment on the results of parts (a) and (b) in light 
of the implicit function theorem (Theorem 6.5). 

44. (a) Consider the family of level sets of the function 

F(x, y) = xy + 1. Use the implicit function theo- 
rem to identify which level sets of this family are 
actually unions of smooth curves in R 2 (i.e., locally 
graphs of C 1 functions of a single variable). 

(b) Now consider the family of level sets of 
F(x, y, z) = xyz + 1. Which level sets of this 
family are unions of smooth surfaces in R 3 ? 

45. Suppose that F(u, v) is of class C 1 and is such that 
F(-2, 1) = 0 and F„(-2, 1) = 7, F„(-2, 1) = 5. Let 
G(x, y, z) = F(x 3 - 2y 2 + z 5 , xy - x 2 z + 3). 

(a) Check that G(-l, 1, 1) = 0. 

(b) Show that we can solve the equation G(x, y, z) = 
0 for z in terms of x and y (i.e., as z = g(x, y), for 
(x,y)near(-l, 1) so that g(-l, 1)= 1). 

46. Can you solve 

X2j2 — x \ cos yi = 5 
X2 sin yi + x\yi = 2 

for yi, y2 as functions of x\, xi near the point 
(x\, X2, yu y2) = (2, 3, n, 1)? What about near the 
point (xi , X2, yi , ya) = (0, 2, n/2, 5/2)? 



Chapter 2 | Differentiation in Several Variables 



x\y\ 



47. Consider the system 

2x 2 y 3 = 1 
xiy{ + x 2 y 2 - 4y 2 yi = -9 . 
x 2 y\ + 3xiyj = 12 

(a) Show that, near the point (x\, x%, y\, y%, y{) = 
(1, 0, —1, 1, 2), it is possible to solve for yi, yj, 
ys in terms of x\, x%. 

(b) From the result of part (a), we may consider y\,y%, 
j3 to be functions of.ti and.*^. Use implicit differ- 

dy\ 

entiation and the chain rule to evaluate (1, 0), 

dx\ 

-i(l,0),and-^(l,0). 
9xi dx\ 

48. Consider the equations that relate cylindrical and 
Cartesian coordinates in R 3 : 

x = r cos 8 
y = r sin 8 . 

z = z 



(a) Near which points of R 3 can we solve for r, 8, and 
z in terms of the Cartesian coordinates? 

(b) Explain the geometry behind your answer in 
part (a). 

49. Recall that the equations relating spherical and Carte- 
sian coordinates in R 3 are 

' x = p sin <p cos 9 

y = p sinip sin#. 

Z = p cos <p 

(a) Near which points of R 3 can we solve for p, <p, and 
8 in terms of x, y, and z? 

(b) Describe the geometry behind your answer in 
part (a). 




Figure 2.75 The tangent line to 
y = f(x) at (xq, f(xo)) crosses the 
x-axis at x = x\. 



2.7 Newton's Method (optional) 

When you studied single-variable calculus, you may have learned a method, known 
as Newton's method (or the Newton-Raphson method), for approximating the 
solution to an equation of the form f(x) = 0, where /: X C R —> R is a differ- 
entiable function. Here's a reminder of how the method works. 

We wish to find a number r such that f(r ) = 0. To approximate r, we make 
an initial guess xq for r and, in general, we expect to find that f(xo) ^ 0. So next 
we look at the tangent line to the graph of / at (x 0 , f(x 0 )). (See Figure 2.75.) 
Since the tangent line approximates the graph of / near (xo, f(xo)), we can 
find where the tangent line crosses the x-axis. The crossing point (jci, 0) will 
generally be closer to (r, 0) than (xq, 0) is, so we take x\ as a revised and improved 
approximation to the root r of f{x) = 0. 

To find x\ , we begin with the equation of the tangent line 

y = f(*o) + /'C*o)(* - *o), 
then set y = 0 to find where this line crosses the x -axis. Thus, we solve the equation 

/(*o) + f'(x 0 )(xi - x 0 ) = 0 

for x\ to find that 

f(xo) 

X] — Xr> . 

f'(x 0 ) 

Once we have x\, we can start the process again using x\ in place of x 0 and 
produce what we hope will be an even better approximation x^ via the formula 

fix,) 

Xi = X\ . 

f'(xi) 

Indeed, we may iterate this process and define Xk recursively by 

f{Xk-\) 



xt = Xk-i k = 1,2,... 

and thereby produce a sequence of numbers xq, x\, . . . , Xk, 



(1) 



2.7 | Newton's Method (optional) 177 



It is not always the case that the sequence {xk\ converges. However, when 
it does, it must converge to a root of the equation f(x) = 0. To see this, let 
L = lim^oo Xk. Then we also have lim^oo jc&_i = L. Taking limits in formula 
(1), we find 

L=L-m, 

/'(!)' 

which immediately implies that f(L) = 0. Hence, L is a root of the equation. 

Now that we have some understanding of derivatives in the multivariable 
case, we turn to the generalization of Newton's method for solving systems of n 
equations in n unknowns. We may write such a system as 

fi(xi, ...,x») = 0 

fi(x\, x„) = 0 

. ■ (2) 

f„(x\, ...,*„) = 0 

We consider the map f: X c R" -> R" defined as f(x) = (/i(x), . . . , / n (x)) (i.e., 
f is the map whose component functions come from the equations in (2). The 
domain X of f may be taken to be the set where all the component functions are 
denned.) Then to solve system (2) means to find a vector r = (rj , . . . , r„) such 
that f(r) = 0. To approximate such a vector r, we may, as in the single- variable 
case, make an initial guess xo for what r might be. If f is differentiable, then we 
know that y = f(x) is approximated by the equation 

y = f(x<,) + Of(x 0 )(x - xo). 

(Here we think of f(xo) and the vectors x and xo as n x 1 matrices.) Then we set 
y equal to 0 to find where this approximating function is zero. Thus, we solve the 
matrix equation 

f(x„) + £>f(x 0 )(x 1 - x 0 ) = 0 (3) 

for xi to give a revised approximation to the root r. Evidently (3) is equivalent to 

Df(xo)( Xl - x 0 ) = -f(x 0 ). (4) 

To continue our argument, suppose that Df(xo) is an invertible n x n matrix, 
meaning that there is a second n x n matrix [Df(xo)] _1 with the property that 
[Df^r'Dftxo) = J Df(x 0 )[Df(x 0 )]" 1 = /„, the n x n identity matrix. (See Ex- 
ercises 20 and 30-38 in §1.6.) Then we may multiply equation (4) on the left by 
[Df(x 0 )] _1 to obtain 

7„(X! -x 0 )=-[Df(x 0 )r 1 f(x 0 ). 

Since I„A = A for any n x k matrix A, this last equation implies that 

Xl =x 0 -[£>f(x 0 )r 1 f(x 0 ). (5) 

As we did in the one -variable case of Newton's method we may iterate formula 
(5) to define recursively a sequence {x^} of vectors by 



x t = xn-[Df( X n)r'%-i) (6) 



Chapter 2 | Differentiation in Several Variables 



-3 + 

Figure 2.76 Finding 
the intersection 
points of the circle 



4 and 



the hyperbola 
Ax 2 - y 2 = A in 
Example 1. 



Note the similarity between formulas (1) and (6). Moreover, just as in the case 
of formula (1), although the sequence {xo, Xi, . . . , x k , . . .} may not converge, if 
it does, it must converge to a root of f(x) = 0. (See Exercise 4.) 

EXAMPLE 1 Consider the problem of finding the intersection points of the cir- 
cle x 2 + y 2 = 4 and the hyperbola 4x 2 — y 2 = 4. (See Figure 2.76.) Analytically, 
we seek simultaneous solutions to the two equations 

x 2 + y 2 = 4 and Ax 2 - y 2 = 4, 

or, equivalently, solutions to the system 



x 2 + y 2 - 4 = 0 



Ax 2 - y 2 



0 



(7) 



To use Newton's method, we define a function f : R -> R by f(x , y) = (x + 



y 



A Ax 1 



A) and try to approximate solutions to the vector equation 



f(x, y) = (0, 0). We may begin with any initial guess, say, 



x 0 



Xo 




T 






l 



and then produce successive approximations Xi, x 2 , 
mula (6). In particular, we have 



to a solution using for- 



NotethatdetDf(x, y) 



2x 
8.v 



2y 
-2y 



Df(x,y) = 

-20xy . You may verify (see Exercise 36 in § 1 .6) that 



[Df(x,y)T l = 



1 



-20xy 



-2y -2y 
-8x 2x 



1 

2 

57 



1 
1 



Thus, 



~x k ~ 
















~Xk-\ 






yk-\_ 






~x k -\ 






yk-i 



[Df(x k _ uyk-i )] f(*k- 1 . yk-i) 
l l 



2 



L 5y k _ 



1 

10%-i J 



4-i + yLi-4 

^U-yU-A 



J x k -\ 



L I0y k -i J 



Xk-l 

yk-\ - 



JX k -\ 



lOxjt i 
10 W -i J 



2.7 | Newton's Method (optional) 179 



Beginning with xo = yo = 1 , we have 

5 ■ l 2 — 8 5 • l 2 — 12 

x x = \ = 1.3 yi = 1 = 1.7 

10-1 10-1 

5(1.3) 2 -8 5(1.7) 2 - 12 

x 2 = 1.3 = 1.265385 y 2 = 1.7 

10(1.3) y 10(1.7) 

= 1.555882, etc. 

It is also easy to hand off the details of the computation to a calculator or a 
computer. One finds the following results: 



k 


x k 


y* 


0 


1 


l 


1 


1.3 


1.7 


2 


1.26538462 


1.55588235 


3 


1.26491115 


1.54920772 


4 


1.26491106 


1.54919334 


5 


1.26491106 


1.54919334 



Thus, it appears that, to eight decimal places, an intersection point of the curves 
is (1.26491106, 1.54919334). 

In this particular example, it is not difficult to find the solutions to (7) exactly. 
We add the two equations in (7) to obtain 



5x z - 8 = 0 



x 2 = 



Thus, x = ±V8/5. If we substitute these values for x into the first equation of 
(7), we obtain 



I + r ~ 4 = 0 y 

Hence, y = ±y/l2/5. Therefore, the four intersection points are 



11 

5 ■ 



Since 78/5 « 1.264911064 and ^/T2/5 « 1.54919334, we see that Newton's 
method provided us with an accurate approximate solution very quickly. ♦ 

EXAMPLE 2 We use Newton's method to find solutions to the system 



x 3 - 5x 2 + 2x - y + 13 = 0 
x 3 +x 2 - 14x - y - 19 = 0 



(8) 



As in the previous example, we define f : R 2 
y + 13, x 3 + x 2 - 14x - y - 19). Then 



R 2 byf(x, y) = (x 3 - 5x 2 + 2x 



Df(x,y) = 



3x 2 - lOx + 2 
3x 2 + 2x - 14 



1 80 Chapter 2 | Differentiation in Several Variables 



so that det Di(x, y) = \2x — 16 and 



1 



1 



[Df(x,y)Y 



Thus, formula (6) becomes 



1 



I2x - 16 
-3x 2 - 2x + 14 
12x - 16 



x k 




~x k -\ 









\2x - 16 
-3x 2 -lOx + 2 
\2x - 16 



1 



12x t . 



-3x 



16 

2*fc_i + 14 



-3x 



12x t . 

2 



i - 16 

lOjCjt-i +2 



12x^-1 — 16 
5x?_, + 2xk-\ - 



12xt_ 



<r-l 



16 



+ X 



k-1 



I4x k - 



y*-i + 13 
y*-i - 19 



6r 2 



I6xk-i — 32 



3.v 



16.T 



\Ax\_ x 



\2xk-\ — 16 

82x t _i - 8y*_i + 6jCjt_iyjt_i + 72 



6^yt-l — 8 



This is the formula we iterate to obtain approximate solutions to (8). 

If we begin with x 0 = (x 0 , y 0 ) = (8, 10), then the successive approximations 
\k quickly converge to (4, 5), as demonstrated in the table below. 



k 


Xk 


yk 


0 


8 


10 


1 


5.2 


-98.2 


2 


4.1862069 


-2.7412414 


3 


4.00607686 


4.82161865 


4 


4.00000691 


4.99981073 


5 


4.00000000 


5.00000000 


6 


4.00000000 


5.00000000 



If we begin instead with x 0 = (50, 60), then convergence is, as you might predict, 
somewhat slower (although still quite rapid): 



k 


Xk 


yk 


0 


50 


60 


1 


25.739726 


-57257.438 


2 


13.682211 


-7080.8238 


3 


7.79569757 


-846.58548 


4 


5.11470969 


-86.660453 


5 


4.1643023 


-1.6486813 


6 


4.00476785 


4.86119425 


7 


4.00000425 


4.99988349 


8 


4.00000000 


5.00000000 


9 


4.00000000 


5.00000000 



2.7 I Exercises 181 



On the other hand, if we begin with xo = (—2, 12), then the sequence of points 
generated converges to a different solution, namely, (—4/3, —25/27): 



k 


Xk 


yk 


0 


-2 


12 


1 


-1.4 


1.4 


2 


-1.3341463 


-0.903122 


3 


-1.3333335 


-0.9259225 


4 


-1.3333333 


-0.9259259 


5 


-1.3333333 


-0.9259259 



In fact, when a system of equations has multiple solutions, it is not always 
easy to predict to which solution a given starting vector xo will converge under 
Newton's method (if, indeed, there is convergence at all). ♦ 

Finally, we make two remarks. First, if at any stage of the iteration process the 
matrix Df(xk) fails to be invertible (i.e., [Df(x,t)] _1 does not exist), then formula 
(6) cannot be used. One way to salvage the situation is to make a different choice 
of initial vector xo in the hope that the sequence {x^} that it generates will not 
involve any noninvertible matrices. Second, we note that if, at any stage, x,t is 
exactly a root of f(x) = 0, then formula (6) will not change it. (See Exercise 7). 



2.7 Exercises 



1 . Use Newton's method with initial vector xo = (1 , — 1) 
to approximate the real solution to the system 

y V = 3 

2ye x + 10y 4 = 0 ' 

2. In this problem, you will use Newton's method to 
estimate the locations of the points of intersection 
of the ellipses having equations 3x 2 + y 2 = 7 and 
x 2 +4y 2 = 8. 

(a) Graph the ellipses and use your graph to give a very 
rough estimate (xo, yo) of the point of intersection 
that lies in the first quadrant. 

(b) Denote the exact point of intersection in the first 
quadrant by (X, Y). Without solving, argue that 
the other points of intersection must be (—X, Y), 
(X, -Y), and (—X, —Y). 

(c) Now use Newton's method with your estimate 
(xq, yo) in part (a) to approximate the first quadrant 
intersection point (X, Y). 

(d) Solve for the intersection points exactly, and com- 
pare your answer with your approximations. 

3. This problem concerns the determination of the points 
of intersection of the two curves with equations jc 3 — 
4y 3 = 1 and x 2 + 4y 2 = 2. 



(a) Graph the curves and use your graph to give rough 
estimates for the points of intersection. 

(b) Now use Newton's method with different initial 
estimates to approximate the intersection points. 

4. Consider the sequence of vectors xo, Xi, . . . , where, 
for k > 1, the vector x* is defined by the Newton's 
method recursion formula (6) given an initial "guess" 
xo at a root of the equation f(x) = 0. (Here we as- 
sume that f:XC R" — > R" is a differentiable func- 
tion.) By imitating the argument in the single-variable 
case, show that if the sequence {xk} converges to a vec- 
tor L and Z)f(L) is an invertible matrix, then L must 
satisfy f(L) = 0. 

5. This problem concerns the Newton's method iteration 
in Example 1 . 

(a) Use initial vector xo = (— 1, 1) and calculate the 
successive approximations Xi, X2, X3, etc. To what 
solution of the system of equations (7) do the ap- 
proximations converge? 

(b) Repeat part (a) with xo = (1, — 1). Repeat again 
with xo = (— 1, —1). 

(c) Comment on the results of parts (a) and (b) and 
whether you might have predicted them. Describe 
the results in terms of Figure 2.76. 



1 82 Chapter 2 | Differentiation in Several Variables 



8. Suppose that f:XCR 2 ->R 2 is differentiable and 
that we write f(x, y) = (f(x, y), g(x, y)). Show that 
formula (6) implies that, for k > 1, 

f(x k -i, y k -i)gy(x k - U Vk-l) ~ g(Xk-l, y k -\)f y (xk-\, yk-i) 
f x (x k -i, y k -i)g y {x k -i, jh) - f y (x k -u yk-i)gx(x k -u y k -\) 

g(x k _i,y k _\)f x {x k -\, y k -i) - /fa-i, y k -i)g x (x k - U y k -i) 
fx(x k -u y k -i)g y (x k -u yk-i) - f y (x k -u yk-\)gx(x k -u yk-i) 



6. Consider the Newton's method iteration in Example 2. 

(a) Use initial vector xo = (1.4, 10) and calculate the 
successive approximations xi, X2, X3, etc. To what 
solution of the system of equations (8) do the ap- 
proximations converge? 

^ (b) Repeat part (a) with x 0 = (1.3, 10). 

(c) In Example 2 we saw that (4, 5) was a solution of 
the given system of equations. Is (1.3, 10)closerto 
(4, 5) or to the limiting point of the sequence you 
calculated in part (b)? 

(d) Comment on your observations in part (c). What 
do these observations suggest about how easily you 
can use the initial vector xo to predict the value of 
lim^oo x k (assuming that the limit exists)? 

7. Suppose that at some stage in the Newton's method it- 
eration using formula (6), we obtain a vector x k that is 
an exact solution to the system of equations (2). Show 
that all the subsequent vectors x^+i, Xk+2, ■ • ■ are equal 
to X£. Hence, if we happen to obtain an exact root via 
Newton's method, we will retain it. 



9. As we will see in Chapter 4, when looking for maxima 
and minima of a differentiable function F:XCR"^ 
R, we need to find the points where DF{x\ , . . . , x n ) = 
[0 • • • 0], called critical points of F. Let F(x, y) = 
4 sin (xy) + x 3 + y 3 . Use Newton's method to approx- 
imate the critical point that lies near (x , y) = (—1, —1). 

1 0. Consider the problem of finding the intersection points 
of the sphere x 2 + y 2 + 7} = 4, the circular cylinder 
x 2 + y 2 = 1, and the elliptical cylinder 4y 2 + z 2 = 4. 

(a) Use Newton's method to find one of the intersec- 
tion points. By choosing a different initial vector 
xo = (xq, yo, zo), approximate a second intersec- 
tion point. (Note: You may wish to use a computer 
algebra system to determine appropriate inverse 
matrices.) 

(b) Find all the intersection points exactly by means of 
algebra and compare with your results in part (a). 



True/False Exercises for Chapter 2 



1 . The component functions of a vector- valued function 8. 
are vectors. 



2. The domain of f(x, y) = |^x + y + 1 
{(x,y)GR 2 | y /0,x # v}. 



3 x 



3. The range of f(x, y) = yx 2 + y L + 1 
{(u, v, w) G R 3 | u > 1}. 



x + y y 



3 x 



x + y y 



4. The function f: R 3 - {(0, 0, 0)} -> R 3 , f(x) = 2x/||x|| 
is one-one. 

5. The graph of x = 9y 2 + z 2 /4 is a paraboloid. 

6. The graph of z + x 2 = y 2 is a hyperboloid. 



The graph of any function of two variables is a level set of 
a function of three variables. 

The level set of any function of three variables is the graph 
of a function of two variables. 



10. 



lim 



2v 2 



(x,v)^(0,0) x 2 + y 2 



1. 



11. If/(x,y) 



4 4 

y — x 
x 2 + y 2 

2 when (x,y) = (0,0) 



when(x,y) /( 0,0\ then / is 



continuous. 



12. 



7. The level set of a function f(x,y,z) is either empty 13. 
or a surface. 



If f(x,y) approaches a number L as (x, y) — > (a, b) 
along all lines through (a,b), then lim( x y) ^( a j,)f(x,y) 
= L. 

If lim x ^ a f(x) exists and is finite, then f is continuous 
at a. 



Miscellaneous Exercises for Chapter 2 



14. f x (a,b)= lim 



/(x, 5) - /(a, fc) 



15. If f(x, y, z) = siny, then V/(x, y, z) = cosy. 

16. If f:R 3 -* R 4 is differentiable, then Df(x) is a 3 x 4 
matrix. 

17. Iff is differentiable at a, then f is continuous at a. 

18. If f is continuous at a, then f is differentiable at a. 

1 9. If all partial derivatives 3 f/dx\ , . . . , 3// 3x„ of a func- 
tion f(xi , . . . , x n ) exist at a = (ai , . . . , a„), then / is 
differentiable at a. 

20. Iff: R 4 -+ R 5 and g: R 4 -+ R 5 are both differentiable 
at a G R 4 , then D(f- g)(a) = Df(a) - Dg(a). 

21. There's a function / of class C 2 such that 

3/ i 3/ 7 

— = y — 2x and — = y — 3xy . 
ox dy 

22. If the second-order partial derivatives of / exist at 
(a, b), then f xy (a, b) = f yx (a, b). 

23. If w = F(x, y, z) and z = g(x, y) where F and g are 
differentiable, then 

dw dF dF dg 
dx dx dz dx 



24. The tangent plane to z = x 1 /(y + 1) at the point 
(—2, 0, —8) has equation z = 12x + 8y + 16. 

25. The plane tangent to xy/z 2 = 1 at (2, 8, —4) has equa- 
tion 4x + y + 2z = 8. 

26. The plane tangent to the surface x 2 + xye z + y 3 = 
1 at the point (2,-1,0) is parallel to the vector 
3i+5j-3k. 

3/ 



27. Dj/(x,y, Z ): 



28. D. k f(x, y, z) 



dy 



3/ 
dz' 



29. If f(x, y) = sinx cosy and v is a unit vector in R 2 , 
thenO^Dv/^-,-) < — . 

30. If v is a unit vector in R 3 and f(x, y, z) = sinx — 
cosy + sinz, then 

-V3 < D y f(x,y,z) < <&. 



Miscellaneous Exercises for Chapter 2 



1. Letf(x) = (i + k) x x. 

(a) Write the component functions of f. 

(b) Describe the domain and range of f. 



order. Complete the following table by matching each 
function in the table with its graph and plot of its level 
curves. 









Graph 


Level curves 


2. Let f(x) = proj 3i _ 2 j +k x, where x = xi + yj + zk. 


Function 




(uppercase 


(lowercase 


(a) Describe the domain and range of f. 


fix, y) 




letter) 


letter) 


(b) Write the component functions of f. 


f(*,y) = 


1 






x 2 + y 2 + 1 








f(x, y) = 


sin y/x 2 + y 2 






3. Let f(x, y) = ^/xy. 


fix, y) = 


(3y 2 - 2x 2 )e-- r2 - 2 >' 2 






(a) Find the domain and range of /. 


fix, y) = 


3 1 2 

y — 3x y 






(b) Is the domain of / open or closed? Why? 


fix, y) = 


x 2 y 2 e -x*-2? 








fix, y) = 


-x 2 -v 2 

ye y 







4. Letg(x, y) = 

(a) Determine the domain and range of g. 

(b) Is the domain of g open or closed? Why? 

5. Figure 2.77 shows the graphs of six functions /(x, y) 
and plots of the collections of their level curves in some 



6. Consider the function /(x, y) = 2 + ln(x 2 + y 2 ). 

(a) Sketch some level curves of /. Give at least those 
at heights, 0, 1, and 2. (It will probably help if you 
give a few more.) 

(b) Using part (a) or otherwise, give a rough sketch of 
the graph of z = f(x, y). 



1 84 Chapter 2 | Differentiation in Several Variables 




J 


1 


1 






-2-10 1 2 
Figure 2.77 Figures for Exercise 5. 




7. Use polar coordinates to evaluate 

,2 



lim 



yx -y 



(x,y)^{0,0) x 2 + y 2 

8. This problem concerns the function 
i 2xy 

f(x,y)-- 

if(x,y) 



C 2 + V 2 

0 



if(x,y) ^(0,0) 
(0,0) 



(a) Use polar coordinates to describe this function. 

(b) Using the polar coordinate description obtained in 
part (a), give some level curves for this function. 

(c) Prepare a rough sketch of the graph of /. 

(d) Determine lim (A: ,,)^ ( o,o) f(x , y), if it exists. 

(e) Is / continuous? Why or why not? 



9. Let 



xy(xy + x 2 ) 



F(x, y) 



if(x,y)#(0,0) 



x 4 + y 4 

0 if (x, y) = (0, 0) 



10. 



Show that the function g(x ) = F (x , 0) is continuous at 
x = 0. Show that the function h(y) = F(0, y) is con- 
tinuous at y = 0. However, show that F fails to be 
continuous at (0, 0). (Thus, continuity in each variable 
separately does not necessarily imply continuity of the 
function.) 

Suppose /: U c R" R is not defined at a point 
a G R" but is defined for all x near a. In other words, 
the domain U of / includes, for some r > 0, the set 
B r = {x G R" | 0 < || x - a|| < r\. (The set B r is just 
an open ball of radius r centered at a with the point 



Miscellaneous Exercises for Chapter 2 



a deleted.) Then we say lim x ^ a /W = +00 if f(x) 
grows without bound as x -> a. More precisely, this 
means that given any N > 0 (no matter how large), 
there is some S > 0 such that if 0 < ||x — a|| < S (i.e., 
ifx € fi,-), then /(a) > N. 

(a) Using intuitive arguments or the preceding tech- 
nical definition, explain why lim^o l/x 2 = 00. 

(b) Explain why 

2 

lim r = 00. 

(x,y)^(l,3) (x - l) 2 + (y - 3) 2 

(c) Formulate a definition of what it means to say that 

lim /(x) = —00. 

x— *a 

(d) Explain why 

1 -x 

lim — ; = —00. 

(x,j.)^(0,0) xy 4 - y 4 + x 3 - x 2 

Exercises 11—1 7 involve the notion of windchill temperature — 
see Example 7 in §2. 1, and refer to the table of windchill values 
on page 85. 

11. (a) Find the windchill temperature when the air tem- 

perature is 25 °F and the windspeed is 10 mph. 

(b) If the windspeed is 20 mph, what air temperature 
causes a windchill temperature of — 1 5 °F? 

12. (a) If the air temperature is 1 0 °F, estimate (to the near- 

est unit) what windspeed would give a windchill 
temperature of —5 °F. 

(b) Do you think your estimate in part (a) is high or 
low? Why? 

13. At a windspeed of 30 mph and air temperature of 35 °F, 
estimate the rate of change of the windchill tempera- 
ture with respect to air temperature if the windspeed is 
held constant. 

14. At a windspeed of 1 5 mph and air temperature of 25 °F, 
estimate the rate of change of the windchill tempera- 
ture with respect to windspeed. 

15. Windchill tables are constructed from empirically de- 
rived formulas for heat loss from an exposed sur- 
face. Early experimental work of P. A. Siple and C. F. 
Passel, 4 resulted in the following formula: 

W = 91 .4 + (t - 91 .4)(0.474 + 0.304V? - 0.0203s). 



Here W denotes windchill temperature (in degrees 
Fahrenheit), t the air temperature (for t < 9 1 .4 °F), and 
s the windspeed in miles per hour (for s > 4 mph). 5 

(a) Compare your answers in Exercises 1 1 and 12 with 
those computed directly from the Siple formula 
just mentioned. 

(b) Discuss any differences you observe between your 
answers to Exercises 1 1 and 12 and your answers 
to part (a). 

(c) Why is it necessary to take t < 91.4°F and 
s > 4 mph in the Siple formula? (Don't look for 
a purely mathematical reason; think about the 
model.) 

1 6. Recent research led the United States National Weather 
Service to employ a new formula for calculating wind- 
chill values beginning November 1, 2001. In partic- 
ular, the table on page 85 was constructed from the 
formula 

W = 35.74 + 0.621f - 35.75j 016 + 0.4275^ 016 . 

Here, as in the Siple formula of Exercise 15, W de- 
notes windchill temperature (in degrees Fahrenheit), 
t the air temperature (for t < 50 °F), and s the wind- 
speed in miles per hour (for s > 3 mph). 6 Compare 
your answers in Exercises 13 and 14 with those com- 
puted directly from the National Weather Service 
formula above. 

17. In this problem you will compare graphically the two 
windchill formulas given in Exercises 15 and 16. 

(a) If W\ (s, t) denotes the windchill function given by 
the Siple formula in Exercise 15 and 0 the 
windchill function given by the National Weather 
Service formula in Exercise 16, graph the curves 
y = Wi(s, 40) and y = Wi(s, 40) on the same set 
of axes. (Let s vary between 3 and 120 mph.) In 
addition, graph other pairs of curves y = W\(s, to), 
y = W2(s, ?o) for other values of to. Discuss what 
your results tell you about the two windchill 
formulas. 

(b) Now graph pairs of curves y = Wi(sq, t), y = 
W2CS0, t) for various constant values so for wind- 
speed. Discuss your results. 

(c) Finally, graph the surfaces z = W\{s, t) and 
z = W2(s, t) and comment. 



"Measurements of dry atmospheric cooling in sub freezing temperatures," Proc. Amer. Phil. Soc, 89 
(1945), 177-199. 

5 From Bob Rilling, Atmospheric Technology Division, National Center for Atmospheric Research 
(NCAR), "Calculating Windchill Values," February 12, 1996. Found online at http://www.atd.ucar.edu/ 
homes/rilling/wc_formula.html (July 31, 2010). 

6 NOAA, National Weather Service, Office of Climate, Water, and Weather Services, "NWS Wind Chill 
Temperature Index." February 26, 2004. <http://www.nws.noaa.gov/om/ windchill> (July 31, 2010). 



1 86 Chapter 2 | Differentiation in Several Variables 



1 8. Consider the sphere of radius 3 centered at the origin. 
The plane tangent to the sphere at (1, 2, 2) intersects 
the x-axis at a point P. Find the coordinates of P. 

1 9. Show that the plane tangent to a sphere at a point P on 
the sphere is always perpendicular to the vector OP 
from the center O of the sphere to P. (Hint: Locate the 
sphere so its center is at the origin in R 3 .) 

20. The surface z = 3x 2 + |x 3 — ^x 4 — 4y 2 is inter- 
sected by the plane 2x — y = I. The resulting intersec- 
tion is a curve on the surface. Find a set of parametric 
equations for the line tangent to this curve at the point 

21. Consider the cone z 2 = x 2 + y 2 . 

(a) Find an equation of the plane tangent to the cone 
at the point (3, -4, 5). 

(b) Find an equation of the plane tangent to the cone 
at the point (a, b, c). 

(c) Show that every tangent plane to the cone must 
pass through the origin. 



22. Show that the two surfaces 



Si: z = xy and S 2 : z = \x 2 — y 
intersect perpendicularly at the point (2, 1,2). 

23. Consider the surface z = x 2 + Ay 2 . 

(a) Find an equation for the plane that is tangent to the 
surface at the point (1, —1,5). 

(b) Now suppose that the surface is intersected with the 
plane x = 1 . The resulting intersection is a curve 
on the surface (and is a curve in the plane x = 1 
as well). Give a set of parametric equations for the 
line in R 3 that is tangent to this curve at the point 
(1 , — 1 , 5). A rough sketch may help your thinking. 

24. A turtleneck sweater has been washed and is now tum- 
bling in the dryer, along with the rest of the laundry. At 
a particular moment to, the neck of the sweater mea- 
sures 1 8 inches in circumference and 3 inches in length. 
However, the sweater is 100% cotton, so that at to the 
heat of the dryer is causing the neck circumference to 
shrink at a rate of 0.2 in/min, while the twisting and 
tumbling action is causing the length of the neck to 
stretch at the rate of 0.1 in/min. How is the volume V 
of the space inside the neck changing at t = ?o? Is V 
increasing or decreasing at that moment? 

25. A factory generates air pollution each day according 
to the formula 

P(S, T) = 3305 2/3 T 4/5 , 

where S denotes the number of machine stations in 
operation and T denotes the average daily tempera- 
ture. At the moment, 75 stations are in regular use and 
the average daily temperature is 15 °C. If the average 



temperature is rising at the rate of 0.2°C/day and the 
number of stations being used is falling at a rate of 
2 per month, at what rate is the amount of pollution 
changing? (Note: Assume that there are 24 workdays 
per month.) 

26. Economists attempt to quantify how useful or satisfy- 
ing people find goods or services by means of utility 
functions. Suppose that the utility a particular individ- 
ual derives from consuming x ounces of soda per week 
and watching y minutes of television per week is 



u(x, y) = 1 



-0.001.v 2 -0.00005v 2 



Further suppose that she currently drinks 80 oz of soda 
per week and watches 240 min of TV each week. If she 
were to increase her soda consumption by 5 oz/week 
and cut back on her TV viewing by 15 min/week, is 
the utility she derives from these changes increasing 
or decreasing? At what rate? 

27. Suppose that w = x 2 + y 2 + z 2 andx = p cos# s'mcp, 
y = p sin 6 sin cp, z = p cos cp . (Note that the equations 
for x, y, and z in terms of p, cp, and 6 are just the con- 
version relations from spherical to rectangular coordi- 
nates.) 

(a) Use the chain rule to compute dw/dp, dw/d(p, 
and dw/dd. Simplify your answers as much as 
possible. 

(b) Substitute p, (p, and 9 for x, y, and z in the original 
expression for w. Can you explain your answer in 
part (a)? 

28. If w = f (— t -^)' show that 

,3w ,3iu 

x 2 y 2 — = 0. 

3x dy 

(You should assume that / is a differentiable function 
of one variable.) 

29. Let z = g(x, y) be a function of class C 2 , and let 
x = e r cos#, v = e r sin#. 

(a) Use the chain rule to find dz/dr and 3 z/ 9 6 in terms 
of dz/dx and dz/dy. Use your results to solve for 
3z/3x and dz/dy in terms of dz/dr and dz/d0. 

(b) Use part (a) and the product rule to show that 



3 2 z 3 2 z _ _ 2r {dh dh 

3x2 + ^2 - e [ dr 2 + d9 2 



30. (a) Use the function /(x, y) = x y (= e ylnx ) and the 

d 

multivariable chain rule to calculate — (w"). 

du 

(b) Use the multivariable chain rule to calculate 

^((sinO cos '). 
dt 



Miscellaneous Exercises for Chapter 2 



31. Use the function f(x, y, z) = x y andthemultivariable 

d , «, 

chain rule to calculate — (u ). 

du 

32. Suppose that /: R" — » R is a function of class C 2 . The 
Laplacian of /, denoted V 2 /, is defined to be 



V 2 / 



9 2 / a 2 / 
— — -i — — + ■ 

dx} dx 2 



+ 



a 2 / 

dx 2 ' 



When n = 2 or 3, this construction is important when 
studying certain differential equations that model phys- 
ical phenomena, such as the heat or wave equations. 
(See Exercises 28 and 29 of §2.4.) Now suppose that 
/ depends only on the distance x = (x\, . . . , x„) is 
from the origin in R" ; that is, suppose that /(x) = g(r) 
for some function g, where r = ||x|| . Show that for all 
x / 0, the Laplacian is given by 



V 2 / 



g'(r) + g"(r). 



33. (a) Consider a function f(x, y) of class C 4 . Show 
that if we apply the Laplacian operator V 2 = 
d 2 /dx 2 + d 2 /dy 2 twice to /, we obtain 



V 2 (V 2 /) 



-L + 2 - 

dx 4 dx 2 dy 2 



+ 



9V 
3y 4 ' 



(b) Now suppose that / is a function of « variables of 
class C 4 . Show that 



V 2 (V 2 /)=£ 



9 4 / 



, dxfdx 1 - 

',7 = 1 ' J 

Functions that satisfy the partial differential equa- 
tion V 2 (V 2 /) = 0 are called Inharmonic func- 
tions and arise in the theoretical study of elasticity. 

34. Livinia, the housefly, finds herself caught in the oven 
at the point (0, 0, 1). The temperature at points in the 
oven is given by the function 

T(x,y,z)= \0(xe-> 2 +ze~ xl ), 

where the units are in degrees Celsius. 

(a) IfLivinia begins to move toward the point (2, 3, 1), 
at what rate (in deg/cm) does she find the temper- 
ature changing? 

(b) In what direction should she move in order to cool 
off as rapidly as possible? 

(c) Suppose that Livinia can fly at a speed of 3 cm/sec. 
If she moves in the direction of part (b), at what 
(instantaneous) rate (in deg/sec) will she find the 
temperature to be changing? 

35. Consider the surface given in cylindrical coordinates 
by the equation z = r cos 36. 

(a) Describe this surface in Cartesian coordinates, that 
is, asz = f{x,y). 



(b) Is / continuous at the origin? (Hint: Think cylin- 
drical.) 

(c) Find expressions for df /dx and df/dy at points 
other than (0, 0). Give values for df /dx and df/ dy 
at (0, 0) by looking at the partial functions of / 
through (x, 0) and (0, y) and taking one-variable 
limits. 

(d) Show that the directional derivative D u f(0, 0) ex- 
ists for every direction (unit vector) u. (Hint: Think 
in cylindrical coordinates again and note that you 
can specify a direction through the origin in the 
xy-plane by choosing a particular constant value 
for 9.) 

(e) Show directly (by examining the expression for 
df/dy when(jc, y) / (0, 0) and also using part (c)) 
that df/dy is not continuous at (0, 0). 

(f) Sketch the graph of the surface, perhaps using a 
computer to do so. 



36. The partial differential equation 

d 2 u d 2 u d 2 u 
Jx 2 + Jv 2 + Jz 2 ~ C ~dt 2 



d 2 u 



is known as the wave equation. It models the motion 
of a wave u ( x , y , z , t ) in R 3 and was originally derived 
by Johann Bernoulli in 1727. In this equation, c is a 
positive constant, the variables x, y, and z represent 
spatial coordinates, and the variable t represents time. 

(a) Let u = cos(x — t) + sin(x + t) — 2e z+l — (y — 
f) 3 • Show that u satisfies the wave equation with 
c= 1. 

(b) More generally, show that if f\, fa, g\, g%, hi, and 
hi are any twice differentiable functions of a single 
variable, then 

u(x, y, z, t) = fi(x - t) + f 2 (x + t) 

+ gi(y - 1) + g 2 (y + 1) 

+ h l (z-t) + h 2 (z + t) 

satisfies the wave equation with c = 1 . 

Let X be an open set in R". A function F: X — > Ris said to be 
homogeneous of degree d if, for all x = (x\, X2, ■ ■ ■ , x n ) e X 
and all t € R such that tx € X, we have 

F(tx\ , tx 2 , . ■ ■ , tx„) = t d F(x\ , x 2 , . . . , x n ). 

Exercises 37—44 concern homogeneous functions. 

In Exercises 37-41, which of the given functions are homoge- 
neous ? For those that are, indicate the degree dof homogeneity. 

37. F(x,y) = x 3 +xy 2 - 6y 3 

38. F(x, y, z) = x 3 y - x 2 z 2 + z & 

39. F(x, y, z) = zy 2 - x 3 + x 2 z 



1 88 Chapter 2 | Differentiation in Several Variables 



40. F(x, y) = e 

41. F(x,y,z) 



JC 3 + X 2 v 



yz'- 



x\ 



z + Ixz 2 



42. If F(x, y, z) is a polynomial, characterize what it 
means to say that F is homogeneous of degree d (i.e., 
explain what must be true about the polynomial if it is 
to be homogeneous of degree d). 

43. Suppose F{x\, X2, . . . , x n ) is differentiable and homo- 
geneous of degree d. Prove Euler's formula: 

dF dF dF 
xi- \-x 2 - 1 \-x„ — =dF. 

OXl 0X2 OX n 



(Hint: Take the equation F(tx\,tX2, ■ ■ ■ ,tx n ) = 
t d F(x\ , X2, . .. ,x n ) that defines homogeneity and dif- 
ferentiate with respect to / .) 

44. Generalize Euler's formula as follows: If F is of class 
C 2 and homogeneous of degree d, then 

d 2 F 
f-r 1 . dxjdx: 

Can you conjecture what an analogous formula 
involving the fcth-order partial derivatives should look 
like? 




3.1 Parametrized Curves and 
Kepler's Laws 

3.2 Arclength and Differential 
Geometry 

3.3 Vector Fields: An 
Introduction 

3.4 Gradient, Divergence, Curl, 
and the Del Operator 

True/False Exercises for 
Chapter 3 

Miscellaneous Exercises 
for Chapter 3 



Z 




y 



X 



Figure 3.1 The path x of 
Example 1. 



Vector-Valued 
Functions 



Introduction 

The primary focus of Chapter 2 was on scalar-valued functions, although general 
mappings from R" to R m were considered occasionally. This chapter concerns 
vector- valued functions of two special types: 

1. Continuous mappings of one variable (i.e., functions x: / C R -+ R", where 
/ is an interval, called paths in R"). 

2. Mappings from (subsets of) R" to itself (called vector fields). 

An understanding of both concepts is required later, when we discuss line and 
surface integrals. 

3.1 Parametrized Curves and Kepler's Laws 

Paths in R n 

We begin with a simple definition. Let / denote any interval in R. (So / can be 
of the form [a, b], (a, b), [a, b), {a, b], [a, oo), (a, oo), (— oo, b], (—00, b), or 
(—00, 00) = R.) 



DEFINITION 1.1 A path in R" is a continuous function x: / -» R" . If / = 

[a, b] for some numbers a <b, then the points x(a) and x(b) are called the 
endpoints of the path x. (Similar definitions apply if / = [a, b), [a, 00), etc.) 



EXAMPLE 1 Let a and b be vectors in R with a / 0. Then the function 
x: (—00, 00) -> R 3 given by 

x(f) = b + ta 

defines the path along the straight line parallel to a and passing through the end- 
point of the position vector of b as in Figure 3.1. (See formula ( 1 ) of § 1 .2.) ♦ 

EXAMPLE 2 The path y: [0, 2tt) -> R 2 given by 

y(f) = (3 cosf, 3 sinf) 

can be thought of as the path of a particle that travels once, counterclockwise, 
around a circle of radius 3 (Figure 3.2). ♦ 



190 Chapter 3 I Vector-Valued Functions 





Figure 3.4 The path x and its 
velocity vector v. 



EXAMPLE 3 The map z: R ->■ R 3 defined by 

z(f) = (a cos t, a sin?, bt), a, b constants {a > 0) 

is called a circular helix, so named because its projection in the xy-plane is a 
circle of radius a. The helix itself lies in the right circular cylinder x 2 + y 2 = a 2 
(Figure 3.3). The value of b determines how tightly the helix twists. ♦ 

We distinguish between a path x and its range or image set x(7), the latter 
being a curve in R". By definition, a path is a function, a dynamic object (at least 
when we imagine the independent variable / to represent time), whereas a curve 
is a static figure in space. With such a point of view, it is natural for us to consider 
the derivative Dx(t ), which we also write as x'(/ ) or v(/), to be the velocity vector 
of the path. We can readily justify such terminology. Since 

x(t) = (x 1 (t),x 2 (t),...,x n (t)) 



is a function of just one variable, 



v(/) = x'(t) = lim 



x(r + At) - x(t) 
At ' 



Thus, v(/) is the instantaneous rate of change of position x(t) with respect to t 
(time), so it can appropriately be called velocity. Figure 3.4 provides an indication 
as to why we draw v(f) as a vector tangent to the path at x(f). Continuing in this 
vein, we introduce the following terminology: 



DEFINITION 1 .2 Let x: / ->■ R" be a differentiable path. Then the velocity 
\(t) = x'(t) exists, and we define the speed of x to be the magnitude of 
velocity; that is, 

Speed = ||v(OI|. 

If v is itself differentiable, then we call V(t) = x"(t) the acceleration of x 
and denote it by m(t). 



EXAMPLE 4 The helix x(t ) = (a cos t , a sin t , bt) has 

v(/) = — a sinf i + a cost j + fok and a(f) = — a cos/ i — a sin/ j. 



3.1 ; Parametrized Curves and Kepler's Laws 



Thus, the acceleration vector is parallel to the xy-plane (i.e., is horizontal). The 
speed of this helical path is 

||v(OII = -vA-a sin t) 2 + (a cos tf + b 2 = y/a 2 + b 2 , 
which is constant. ♦ 

The velocity vector v is important for another reason, namely, for finding equations 
of tangent lines to paths. The tangent line to a differentiate path x, at the point 
x 0 = x(?o), is the line through xo that is parallel to any (nonzero) tangent vector 
to x at xo. Since v(t), when nonzero, is always tangent to x(t), we may use equa- 
tion ( 1 ) of § 1 .2 to obtain the following vector parametric equation for the tangent 
line: 

l(s) = Xo + SV(). (1) 

Here vo = v(to) and s may be any real number. 

In equation (1), we have 1(0) = Xo. To relate the new parameter s to the 
original parameter t for the path, we set s = t — to and establish the following 
result: 



PROPOSITION 1 .3 Let x be a differentiate path and assume that v 0 = 
v(?o) ^ 0. Then a vector parametric equation for the line tangent to x at x 0 = x(t 0 ) 
is either 

l(s) = x 0 + s\ 0 (2) 

or 

1(0 = x 0 + (t - f 0 )v 0 . (3) 

(See Figure 3.5.) 



EXAMPLE 5 If x(?) = (it + 2, t 2 - 7, t - t 2 ), we find parametric equations 
for the line tangent to x at (5, — 6, 0) = x(l). 

For this path, v(0 = x'(t) = 3i + 2/j + (1 - 2t)k, so that 



Thus, by formula (3), 

1(0 



v 0 = v(l) = 3i + 2j-k. 



(5i-6j) + (f- l)(3i + 2j-k). 



Taking components, we read off the parametric equations for the coordinates 
of the tangent line as 



x = 3t +2 
y = 2t - 8 . 
z=\-t 



The physical significance of the tangent line is this: Suppose a particle of 
mass m travels along a path x. If, suddenly, at t = to, all forces cease to act on the 
particle (so that, by Newton's second law of motion F = ma, we have a(/) = 0 
for t > to), then the particle will follow the tangent line path of equation (3). 



192 Chapter 3 I Vector-Valued Functions 



EXAMPLE 6 If Roger Ramjet is fired from a camion, then we can use vectors 
to describe his trajectory. (See Figure 3.6.) 

y 



Roger's path 




x 



Figure 3.6 Roger Ramjet's path. 

We'll assume that Roger is given an initial velocity vector vo by virtue of the 
firing of the cannon and that thereafter the only force acting on Roger is due to 
gravity (so, in particular, we neglect any air resistance). Let us choose coordinates 
so that Roger is initially at the origin, and throughout our calculations we '11 neglect 
the height of the cannon. Let x(r) = (x(f), y(t)) denote Roger's path. Then the 
information we have is 

a (t) = x"(t)=-g\ 
(i.e., the acceleration due to gravity is constant and points downward); hence, 

v(0) = x'(0) = v 0 

and 



x(0) = 0. 

Since a(/) = V(t), we simply integrate the expression for acceleration compo- 
nentwise to find the velocity: 



v(0 



= j a(t)dt = j -g]dr = -gtj + 



Here c is an arbitrary constant vector (the "constant of integration"). Since v(0) = 
vo, we must have c = vo, so that 

v(0 = -gtj + v 0 . 
Integrating again to find the path, 

x(t) = j v(t)dt = j (-gt\ + \ 0 )dt = -^gt 2 ] + t\ 0 + d, 

where d is another arbitrary constant vector. From the remaining fact that x(0) = 0, 
we conclude that 

1 7 

x(0 = -^gn + ty 0 (4) 

describes Roger's path. 

To understand equation (4) better, we write vo in terms of its components: 

vo = vo cos 9 i + vo sin 9 j. 

Here vo = ||vo|| is the initial speed. (We're really doing nothing more than 
expressing the rectangular components of vo in terms of polar coordinates. 



3.1 : Parametrized Curves and Kepler's Laws 



See Figure 3.7.) Thus, 



x(r) = — \gt 2 \ + f(i> 0 cos#i + vosin# j) 



= (v 0 cos 9)t\ + I (u 0 sin 9)t — -gt 2 



1 

2 

From this, we may read off the parametric equations: 

x = (vo cos6)t 

1 , - 
y = (v 0 sinfl)/ - -gt z 

from which it is not difficult to check that Roger's path traces a parabola. ♦ 

Here are two practical questions concerning the set-up of Example 6: First, for 
a given initial velocity, how far does Roger travel horizontally? Second for a given 
initial speed how should the cannon be aimed so that Roger travels (horizontally) 
as far as possible? To find the range of the cannon shot and thereby answer the 
first question, we need to know when y = 0 (i.e., when Roger hits the ground). 
Thus, we solve 

(u 0 sin 6>)r - \gt 2 = t(v 0 sm9 - \gt) = 0 

for t. Hence, y = 0 when t = 0 (which is when Roger blasts off) and when 
t = (2vq sin9)/g. At this later time, 

(2vo sin#\ sin 29 
— = ■ (5) 

8 J S 

Formula (5) is Roger's horizontal range for a given initial velocity. To maximize 
the range for a given initial speed vo, we must choose 9 so that (v 2 sin29)/g is 
as large as possible. Clearly, this happens when sin 29 = 1 (i.e., when 9 = jt/4). 



Kepler's Laws of Planetary Motion (optional) 

Since classical antiquity, individuals have sought to understand the motions of 
the planets and stars. The majority of the ancient astronomers, using a combina- 
tion of crude observation and faith, believed all heavenly bodies revolved around 
the earth. Fortunately, the heliocentric (or "sun-centered") theory of Nicholas 
Copernicus (1473-1543) did eventually gain favor as observational techniques 
improved. However, it was still believed that the planets traveled in circular or- 
bits around the sun. This circular orbit theory did not correctly predict planetary 
positions, so astronomers postulated the existence of epicycles, smaller circular 
orbits traveling along the major circular arc, an example of which is shown in 
Figure 3.8. Although positional calculations with epicycles yielded results closer 
to the observed data, they still were not correct. Attempts at further improvements 
were made using second- and third-order epicycles, but any gains in predictive 
power were made at a cost of considerable calculational complexity. A new idea 
was needed. Such inspiration came from Johannes Kepler (1571-1630), son of a 
saloonkeeper and assistant to the Danish astronomer Tycho Brahe. The classical 
astronomers were "stuck on circles" for they believed the circle to be a perfect 
form and that God would use only such perfect figures for planetary motion. 
Kepler, however, considered the other conic sections to be as elegant as the cir- 
cle and so hypothesized the simple theory that planetary orbits are elliptical. 
Empirical evidence bore out this theory. 



194 Chapter 3 | Vector-Valued Functions 




Figure 3.9 Kepler's second law 
of planetary motion: If 
t 2 — t\ = t 4 — f 3 , then A\ = A 2 , 
where A ; and A 2 are the areas of 
the shaded regions. 



Kepler's three laws of planetary motion are 

1. The orbit of a planet is elliptical, with the sun at a focus of the ellipse. 

2. During equal periods of time, a planet sweeps through equal areas with respect 
to the sun. (See Figure 3.9.) 

3. The square of the period of one elliptical orbit is proportional to the cube of 
the length of the semimajor axis of the ellipse. 

Kepler's laws changed the face of astronomy. We emphasize, however, that 
they were discovered empirically, not analytically derived from general physical 
laws. The first analytic derivation is frequently credited to Newton, who claimed 
to have established Kepler's laws (at least the first and third laws) in Book I of 
his Philosophiae Naturalis Principia Mathematica (1687). However, a number of 
scientists and historians of science now consider Newton's proof of Kepler's first 
law to be flawed and that Johann Bernoulli (1667-1748) offered the first rigorous 
derivation in 1710. 1 In the discussion that follows, Newton's law of universal 
gravitation is used to prove all three of Kepler's laws. 

In our work below, we assume that the only physical effects are those be- 
tween the sun and a single planet — the so-called two-body problem. (The n-body 
problem, where n > 3 is, by contrast, an important area of current mathematical 
research.) To set the stage for our calculations, we take the sun to be fixed at the 
origin O in R 3 and the planet to be at the moving position P. We also need the 
following two "vector product rules," whose proofs we leave to you: 



PROPOSITION 1.4 

1. If x and y are differentiable paths in R", then 

d dx dy 
— (x • y) = v • h x • — . 

dt y J dt dt 

2. If x and y are differentiable paths in R 3 , then 

d dx dy 

— (x xy)= — xy + xx — . 
dt y J dt J dt 



First, we establish the following preliminary result: 



PROPOSITION 1 .5 The motion of the planet is planar, and the sun lies in the 
planet's plane of motion. 



PROOF Let r = OP. Then r is a vector whose representative arrow has its tail 
fixed at O. (Note that r = r(t); that is, r is a function of time.) If v = r'(f), we 
will show that r x v is a constant vector c. This result, in turn, implies that r must 
always be perpendicular to c and, hence, that r always lies in a plane with c as 
normal vector. 

To show that r x v is constant, we show that its derivative is zero. By part 2 
of Proposition 1.4, 

d dr dy 

— (r xv)= — xv + rx — = vxv+rxa, 

dt dt dt 



For an indication of the more recent controversy surrounding Newton's mathematical accomplishments, 
see R. Weinstock, "Isaac Newton: Credit where credit won't do," The College Mathematics Journal, 25 
(1994), no. 3, 179-192, andC. Wilson, "Newton's orbit problem: A historian's response," Ibid., 193-200, 
and related papers. 



3.1 i Parametrized Curves and Kepler's Laws 



by the definitions of velocity and acceleration. We know that vxv = 0 (why?), so 

d 

— (r x v) = r x a. (6) 
dt 

Now we use Newton's laws. Newton's law of gravitation tells us that the planet 
is attracted to the sun with a force 

GMm 

F = — u, (7) 

where G is Newton's gravitational constant (= 6.6720 x 10~ n Nm 2 /kg 2 ), M 
is the mass of the sun, m is the mass of the planet (in kilograms), r = ||r||, and 
u = r/ 1| i* || (distances in meters). On the other hand Newton's second law of 
motion states that, for the planet, 

F = ma. 

Thus, 

GMm 
ma = r — u, 

r L 

or 

GM 

a = -r. (8) 

Therefore, a is just a scalar multiple of r and hence is always parallel to r. In 
view of equations (6) and (8), we conclude that 



d 

— (r xv) = rxa = 0 

dt 



(i.e., that r x v is constant). 



THEOREM 1 .6 (Kepler's first law) In a two-body system consisting of one 
sun and one planet, the planet's orbit is an ellipse and the sun lies at one focus of 
that ellipse. 



PROOF We will eventually find a polar equation for the planet's orbit and see 
that this equation defines an ellipse as described. We retain the notation from 
the proof of Proposition 1.5 and take coordinates for R 3 so that the sun is at the 
origin, and the path of the planet lies in the xy-plane. Then the constant vector 
c = r x v used in the proof of Proposition 1.5 may be written as ck, where c is 
some nonzero real number. This set-up is shown in Figure 3.10. 

z 

c = r x v 




Figure 3.10 Establishing Kepler's laws. 



196 Chapter 3 I Vector-Valued Functions 



Step 1. We find another expression for c. By definition of u in formula (7), 
r = ru, so that, by the product rule, 



d du dr 

v = — (ru) = r— + — u. 

dt dt dt 



Hence, 



/ du dr \ 2 / du\ dr 
c = r x v = (ru) x r 1 u =r ux — + r — (u x u). 

\ dt dt J \ dt J dt 

Since uxu must be zero, we conclude that 

c = r'(uxf). ,9, 

Step 2. We derive the polar equation for the orbit. Before doing so, however, 
note the following result, whose proof is left to you as an exercise: 



PROPOSITION 1.7 If x(r) has constant length (i.e., ||x(f)|| is constant for 
all r), then x is perpendicular to its derivative dx/dt. 



Continuing now with the main argument, note that the vector r(r) is denned 
so that its magnitude is precisely the polar coordinate r of the planet's position. 
Using equations (8) and (9), we find that 



axe 



GM 



u x r u x 



du 

dt 



= -GM 



= GM 



duX 

U X | U X 

dt ) 



du\ 

U X X u 

dt J 



= GM 



= GM 



(u-u) 



du 

dt 



du 
u- — | u 

dt 



du 
1 Ou 

dt 



(see Exercise 27 of § 1 .4) 
(by Proposition 1.7) 



= - (GMu), 
dt 

since G and M are constant. On the other hand, we can "reverse" the product rule 
to find that 

d\ 

a x c = — x c 
dt 

dy dc 
= — xc + vx— (since c is constant) 
dt dt 



= — (v X c). 

dt 



3.1 ; Parametrized Curves and Kepler's Laws 




Figure 3.1 1 The angle 9 is 
the angle between r and d. 



Thus, 



and, hence, 



d d 
a x c = — (GMu) = — (v x c), 
dt dt 



v x c = GMu + d, 



(10) 

where d is an arbitrary constant vector. Because both v x c and u lie in the xy- 
plane, so must d. 

Let us adjust coordinates, if necessary, so that d points in the i-direction (i.e., 
so that d = di for some d e R). This can be accomplished by rotating the whole 
set-up about the z-axis, which does not lift anything lying in the xy-plane out of 
that plane. Then the angle between r (and hence u) and d is the polar angle 6 as 
shown in Figure 3.11. 

By Theorem 3.3 of Chapter 1, 

u-d= ||u| 

Since c = ||c||, 

2 



Idll cos# = dcosi 



(11) 



c • c 

= (r x v) • c 
= r • (v x c) 
= ru-(GMu + d) 



Hence, 



(Why? See formula (4) of §1.4.) 
by equation (10). 

GMr + rd cos6» 



by equation (11). We can readily solve this equation for r to obtain 

c 2 

r = 



(12) 



GM + d cosO 
the polar equation for the planet's orbit. 

Step 3. We now check that equation (12) really does define an ellipse by 
converting to Cartesian coordinates. First, we'll rewrite the equation as 

c 2 (c 2 /GM) 
GM + d cosO ~ l + (d/GM)cose' 

and then let p = c 2 / GM, e = d/ GM for convenience. (Note that p > 0.) Hence, 
equation (12) becomes 

r = ■ (13) 

1 + e cos e 

A little algebra provides the equivalent equation, 

r = p — er cos6. (14) 

Now r cose = x (x being the usual Cartesian coordinate), so that equation (14) 
is equivalent to 

r = p — ex. 

To complete the conversion, we square both sides and find, by virtue of the fact 
that r 2 = x 2 + y 2 , 



x 2 + y 2 = p 2 



2pex + e 2 x 2 . 



198 Chapter 3 I Vector-Valued Functions 



A little more algebra reveals that 

(1- e 2 )x 2 + 2pex + y 2 = p 2 . (15) 

Therefore, the curve described by the preceding equation is an ellipse if 0 < 
\e\ < 1, a parabola if e = ±1, and a hyperbola if \e \ > 1 . Analytically, there is no 
way to eliminate the last two possibilities. Indeed, "uncaptured" objects such as 
comets or expendable deep space probes can have hyperbolic or parabolic orbits. 
However, to have a closed orbit (so that the planet repeats its transit across the 
sky), we are forced to conclude that the orbit must be elliptical. 

More can be said about the elliptical orbit. Dividing equation (15) by 1 — e 2 
and completing the square in x, we have 

\ 2 2 2 

x + 



1 -e 2 (1 -e 2 ) 2 ' 
This is equivalent to the rather awkward-looking equation 

(x + pe/(l-e 2 )f y 2 

p 2 /(l - e 2 ) 2 p 2 /{\ - e 2 ) y J 

From equation (16), we see that the ellipse is centered at the point (— pe/(l — e 2 ), 
0), that its semimajor axis has length a = p/(l — e 2 ), and that its semiminor axis 
has length b = p/^/l — e 2 . The foci of the ellipse are at a distance 



yV - b 2 



P 2 P 2 P\e\ 



(1 - e 2 ) 2 1 - e 2 



from the center. (See Figure 3.12.) Hence, we see that one focus must be at the 
origin, the location of the sun. Our proof is, therefore, complete. ■ 

Fortunately, all the toil involved in proving the first law will pay off in proofs 
of the second and third laws, which are considerably shorter. Again, we retain all 
the notation we already introduced. 



THEOREM 1 .8 (Kepler's second law) During equal intervals of time, aplanet 
sweeps through equal areas with respect to the sun. 




Figure 3.1 2 The ellipse of equation (16). 



3.1 ; Parametrized Curves and Kepler's Laws 



Po(r 0 , B 0 ) 



P(r, 6) 




Figure 3.1 3 The shaded area A(9) is 
given by \r 2 dip. 



PROOF Fix one point Pq on the planet's orbit. Then the area A swept between 
Po and a second (moving) point P on the orbit is given by the polar area integral 



A{6) 



f 6 1 2 

L 2 r 



dip. 



(See Figure 3.13.) Thus, we may reformulate Kepler's law to say that dA/dt is 
constant. We establish this reformulation by relating dA/dt to a known constant, 
namely, the vector c = r x v. 

By the chain rule (in one variable), 

dA _ dA d9 

dt d6 dt ' 

By the fundamental theorem of calculus, 

dA d f 9 1 , , 1 r ,~ 



Hence, 



dA _ 1 2 dO 
dt ~ 2 r dt' 



(17) 



Now, we relate c to dO/dt by means of equation (9). Therefore, we compute 

u x du/dt in terms of 0. Recall that u = - r and r = r cos 0 i + r sin 6* j. Thus, 

r 

cos 0 i + sin 0 j 



u 

du 

dt dt ' dt 

Hence, it follows by direct calculation of the cross product that 



dO dO 
smtf — i + cosO — j. 



c = r u x 



du 

dt 



2 d0 
= r — k, 
dt 



so c = ||c|| = r 2 d0/dt, and equation (17) implies that 

dA _ 1 

~dt ~ 2 C ' 

a constant. 



(18) 



THEOREM 1 .9 (Kepler's third law) If T is the length of time for one plane- 
tary orbit, and a is the length of the semimajor axis of this orbit, then T 2 = Ka 3 
for some constant K. 



Chapter 3 | Vector-Valued Functions 



3.1 Exercises 



PROOF We focus on the total area enclosed by the elliptical orbit. The area of an 
ellipse whose semimajor and semiminor axes have lengths a and b, respectively, 
is nab. This area must also be that swept by the planet in the time interval [0, T]. 
Thus, we have 

' T dA 
— dt 
dt 

T 1 

-cdt by equation ( 1 8) 



nab = I 

Jo 

-L 



Hence, 



1 

= 2 CT - 

T = 2 -^, so r 2 = 4 " W . (19) 

c c 2 

Now, b and c are related to a, so these quantities must be replaced before we are 
done. In particular, from equation (16), b 2 = p 2 /(l — e 2 ), so 



b 2 = pa. 



Also 

P 



c 2 



GM 

(See equations (12) and (13).) With these substitutions, the result in (19) becomes 
t2 = An 2 a 2 {pa) = /4^r 2 \ ^ 



pGM \Gm) 



This last equation shows that T 2 is proportional to a 3 , but it says even more: 
The constant of proportionality 4n 2 /GM depends entirely on the mass of the 
sun — the constant is the same for any planet that might revolve around the sun. 



In Exercises 1-6, sketch the images of the following paths, us- 
ing arrows to indicate the direction in which the parameter 
increases: 

. \x = It - 1 

1. { , , , -1 < t < 1 
[y = 3 - t 

2. x(t) = e' i + <?"' j 

_ \x = t cos t , , 

3. { . , -6n < t < 6jt 
I y = t sin t ~ ~ 

. \x = 3 cost „ - 
I y = 2 sin 2t ~ ~ 

5. x(r) = (f,3r 2 +l,0) 

6. x(t) = (t, t 2 , t 3 ) 

Calculate the velocity, speed, and acceleration of the paths 
given in Exercises 7—10. 

7. x(r) = (3f-5)i + (2? + 7)j 



8. x(f ) = 5 cos t i + 3 sin t j 

9. x(r) = (t sin*, t cos t, t 2 ) 

10. x(?) = (e',e 2 ',2e') 

In Exercises 11—14, (a) use a computer to give a plot of the 
given path x over the indicated interval for t; identify the di- 
rection in which t increases, (b) Show that the path lies on the 
given surface S. 

^ 11. x(?) = (3cosjrf,4sin7r?,2?), -4 < t < 4; S is ellip- 

x 2 y 2 

tical cylinder \- - — = 1 . 

9 16 

^12. x(f) = (t cost, t shU, t), -20 < t < 20; S is cone 

Z 2 = x 2 + y 2 . 

^ 13. x(?) = (t sin2r, t cos2r, t 2 ), -6 < t < 6; S is para- 
boloid z = x 2 + y 2 . 



3.1 | Exercises 201 



^ 14. x(r) = (2cosf, 2sinr, 3 sin8f), 0 < t < 2jt; S is cy- 
linders 2 + y 2 = 4. 

In Exercises 1 5-18, find an equation for the line tangent to the 
given path at the indicated value for the parameter. 

15. x(r) = te~> i + e 3 ' j, t = 0 

16. x(t) = 4cos?i— 3sinr j + 5tk, t = tt/3 

17. x(t) = (t 2 ,t\t 5 ), t = 2 

18. x(r) = (cos(e'), 3 - t 2 , t), t = 1 

19. (a) Sketch the path x(f) = (t, f 3 - It + 1). 

(b) Calculate the line tangent to x when t = 2. 

(c) Describe the image of x by an equation of the form 
y = f(x) by eliminating t . 

(d) Verify your answer in part (b) by recalculating the 
tangent line, using your result in part (c). 

Exercises 20-23 concern Roger Ramjet and his trajectory when 
he is shot from a cannon as in Example 6 of this section. 

20. Verify that Roger Ramjet's path in Example 6 is indeed 
a parabola. 

21 . Suppose that Roger is fired from the cannon with an 
angle of inclination 6 of 60° and an initial speed Vq of 
100 ft/sec. What is the maximum height Roger attains? 

22. Suppose that Roger is fired from the cannon with an an- 
gle of inclination 6 of 60° and that he hits the ground 
1/2 mile from the cannon. What, then, was Roger's 
initial speed? 

23. If Roger is fired from the cannon with an initial speed of 
250 ft/sec, what angle of inclination 6 should be used 
so that Roger hits the ground 1 500 ft from the cannon? 

24. Gertrude is aiming a Super Drencher water pistol at 
Egbert, who is 1.6 m tall and is standing 5 m away. 
Gertrude holds the water gun 1 m above ground at an 
angle a of elevation. (See Figure 3.14.) 

(a) If the water pistol fires with an initial speed of 
7 m/sec and an elevation angle of 45°, does Egbert 
get wet? 



(b) If the water pistol fires with an initial speed of 
8 m/sec, what possible angles of elevation will 
cause Egbert to get wet? (Note: You will want to 
use a computer algebra system or a graphics cal- 
culator for this part.) 

25. A malfunctioning rocket is traveling according to the 
pathx(r) = (e 2 ', 3f 3 — 2t, t - j) in the hope of reach- 
ing a repair station at the point (7e 4 , 35, 5). (Here 
t represents time in minutes and spatial coordinates 
are measured in miles.) Alt = 2, the rocket's engines 
suddenly cease. Will the rocket coast into the repair 
station? 

26. Two billiard balls are moving on a (coordina- 
tized) pool table according to the respective paths 



x(f) = (t 



!, j - 1J and y(f) = (t, 5 - r), where 
t represents time measured in seconds. 

(a) When and where do the balls collide? 

(b) What is the angle formed by the paths of the balls 
at the collision point? 

27. Establish part 1 of Proposition 1.4 in this section: If x 
and y are differentiable paths in R" , show that 



dt 



(x-y): 



dx 



dy 
dt ' 



28. Establish part 2 of Proposition 1.4 in this section: If x 
and y are differentiable paths in R 3 , show that 



dt 



dx dy 
(xxy) = — xy + xx — . 

dt at 



29. Prove Proposition 1.7. 

30. (a) Show that the path x(f) = (cos t, cos t sin t, sin 2 t) 

lies on a unit sphere. 

(b) Verify that x(r) is always perpendicular to the ve- 
locity vector v(f ). 

(c) Use Proposition 1 .7 to show that if a differentiable 
path lies on a sphere centered at the origin, then 
its position vector is always perpendicular to its 
velocity vector. 




Figure 3.14 Figure for Exercise 24. 



Chapter 3 | Vector-Valued Functions 



31. Consider the path 

x = (a + b cos cot ) cos t 
■ y = (a + b cos cot) sin t , 
z = b sin cot 

where a, b, and co are positive constants and a > b. 

(a) Use a computer to plot this path when 

i. a = 3, b = 1, and co = 15. 

ii. a = 5, b = 1, and co = 15. 

iii. a = 5, b = 1, and w = 25. 

Comment on how the values of a, b, and <w affect 
the shapes of the image curves. 

(b) Show that the image curve lies on the torus 

(vV + y 2 -a) 2 + z 2 = b 2 . 

(A torus is the surface of a doughnut.) 

32. For the path x(f ) = (e' cos t, e' sin f ), show that the an- 
gle between x(t) and x'(t) remains constant. What is 
the angle? 

33. Consider the path x: R R 2 , x(f ) = (t 2 , t 3 - t). 

(a) Show that this path intersects itself, that is, that 
there are numbers t\ and ti such that x(f i ) = x(?2). 

(b) At the point where the path intersects itself, it 
makes sense to say that the image curve has two 
tangent lines. What is the angle between these tan- 
gent lines? 

34. Although the path x : [0, 2tt] — > R 2 , x(f ) = 
(cosf, sinf) may be the most familiar way to give a 
parametric description of a unit circle, in this problem 
you will develop a different set of parametric equations 
that gives the x- and v-coordinates of a point on the 
circle in terms of rational functions of the parameter. 
(This particular parametrization turns out to be useful 
in the branch of mathematics known as number theory.) 

To set things up, begin with the unit circle x 1 + 
y 2 = 1 and consider all lines through the point (—1,0). 
(See Figure 3.15.) Note that every line other than the 



vertical line x = — 1 intersects the circle at a point 
(x, y) other than (—1, 0). Let the parameter t be the 
slope of the line joining (—1,0) and a point (x, y) on 
the circle. 



y 



(-1,0)1 






2" Slope? 








X 



Figure 3.15 Figure for Exercise 34. 



(a) Give an equation for the line of slope t joining 
(—1,0) and (x, y). (Your answer should involve 
x, y, and t.) 

(b) Use your answer in part (a) to write y in terms of 
x and t. Then substitute this expression for y into 
the equation for the unit circle. Solve the resulting 
equations for x in terms of /. Your answer(s) for x 
will give the points of intersection of the line and 
the circle. 

(c) Use your result in part (b) to give a set of paramet- 
ric equations for points (x, y) on the unit circle. 

(d) Does your parametrization in part (c) cover the 
entire circle? Which, if any, points are missed? 

35. Let x(f) be a path of class C 1 that does not pass through 
the origin in R 3 . If x(fo) is the point on the image of x 
closest to the origin and x'(fo) / 0, show that the po- 
sition vector x(?o) is orthogonal to the velocity vector 
x'(fo). 



3.2 Arclength and Differential Geometry 

In this section, we continue our general study of parametrized curves in R 3 , 
considering how to measure such geometric properties as length and curvature. 
This can be done by defining three mutually perpendicular unit vectors that form 
the so-called moving frame specially adapted to a path x. Our study takes us 
briefly into the branch of mathematics called differential geometry, an area where 
calculus and analysis are used to understand the geometry of curves, surfaces, 
and certain higher-dimensional objects (called manifolds). 



3.2 | Arclength and Differential Geometry 



Length of a Path 

For now, let x: [a, b] —> R 3 be a C 1 path in R 3 . Then we can approximate the 
length L of x as follows: First, partition the interval [a, b] into n subintervals. 
That is, choose numbers to, t\, ...,/„ such that a = to < t\ < ■ ■■ < t„ = b. If, 
for i = 1, . . . , n, we let As, denote the distance between the points x(r,_i) and 
x(ti) on the path, then 

n 

1=1 

(See Figure 3.16.) We have x(f) = (x(t), y(t), z(t)), so that the distance formula 
(i.e., the Pythagorean theorem) implies 

As t = J Ax 2 + Ay, 2 + Az 2 , 

where Axj = x(t { ) - x(ti-\), Ay t = y(u) - y(ti-\), and Az,- = z{U) - zft-i). It 
is entirely reasonable to hope that the approximation in (1) improves as the At t 's 
become closer to zero. Hence, we define the length L of x to be 

n 

L= lim Y^A Xi 2 + Ay,- 2 + Az,- 2 . (2) 

max Ar,-^0 ~—f 
i=l 

Now, we find a way to rewrite equation (2) as an integral. On each subinterval 
fj], apply the mean value theorem (three times) to conclude the following: 

1. There must be some number t* in [£,-_i, ?, ] such that 

x(tt) - x(»,_i) = x'(ff)fe - fj_i); 

that is, Ax, = x'{t*)At t . 

2. There must be another number t** in [/, i , r,] such that 

Ay i =y'(t**)At t . 

3. There must be a third number t*** in [/,_!, f, ] such that 

Az, =z'(t***)At h 
Therefore, with a little algebra, equation (2) becomes 

L= lim V Jx'{tff + y'(t**) 2 + z'(t***) 2 At h (3) 

max Ar,^0 z — ■f V 
( = 1 

When the limit appearing in equation (3) is finite, it gives the value of the definite 
integral 

b 

Jx>(t) 2 + y>(t) 2 + z'(t) 2 dt. 

Note that the integrand is precisely ||x'(f)||, the speed of the path. (This makes 
perfect sense, of course. Speed measures the rate of distance traveled per unit 
time, so integrating the speed over the elapsed time interval should give the total 
distance traveled.) Moreover, it's not hard to see how we should go about defining 
the length of a path in R" for arbitrary n. 



L 



204 Chapter 3 | Vector- Valued Functions 



DEFINITION 2.1 The length L(x) of a C 1 path x: [a, b] -> R" is found by 
integrating its speed: 



= f llx' 

J a 



L(x) = / \\x'(t)\\dt. 




Figure 3.17 AC 1 path. 




x(a) 



Figure 3.1 

x: [a, b] 



x(b) 



8 Apiecewise C 1 path 
R 3 . 



EXAMPLE 1 To check our definition in a well-known situation, we compute 
the length of the path 

x: [0, 2n] — > R 2 , x(/) = (a cost, a sin t), a > 0. 



We have 



so 



x'(r) = — a sinr i + a cos? j, 



||x'(f)ll = V o 2 sin 2 1 + a 2 cos 2 f = a. 
Thus, Definition 2.1 gives 



L(x) 



2,t 



: dt = lit a. 



Since the path traces a circle of radius a once, the length integral works out to be 
the circumference of the circle, as it should. ♦ 

EXAMPLE 2 For the helix x(t) = (a cost, a sint, bt), 0 < t < 2ir, we have 

x'(t) = —a sin t i + a cos t j + b k, 
so that ||x'(0|| = Va 2 + £ 2 , and 



L(x) 



= / V<s 2 + b 2 dt = + b 2 . 

Jo 



When b = 0, the helix reverts to a circle and the length integral agrees with the 
previous example. ♦ 

Although we have defined the length integral only for C 1 (or "smooth- 
looking") paths, there is no problem with extending our definition to the piecewise 
C 1 case. By definition, a C 1 path is one with a continuously varying velocity vec- 
tor, and so it typically looks like the path in Figure 3. 1 7. A piecewise C 1 path is one 
that may not be C 1 but instead consists of finitely many C 1 chunks. A continuous, 
piecewise C 1 path that is notC 1 typically looks like the path in Figure 3. 18. Each 
of the three portions of the path defined for (i) a < t < t\, (ii) t\ < t < h, and 
(iii) t2 < t < b is of class C 1 , but the velocity, if nonzero, would be discontinuous 
at t = t\ and t = t2 - To define the length of a piecewise C 1 path, all we need do is 
break up the path into its C 1 pieces, calculate the length of each piece, and add to 
get the total length. For the piecewise C 1 path shown in Figure 3.18, this means we 
would take 

rt\ pt 2 rb 

/ \\x'(t)\\dt+ \\x'(t)\\dt+ ||x'(f)||df 

J a J t\ J t2 

to be the length. 



3.2 | Arclength and Differential Geometry 



Warning Even if a path is continuous, the definite integral in Definition 2.1 
may fail to exist. An example of such an unfortunate situation is furnished by the 
pathx: [0, 1] R 2 , 

1 

t sin 

x(t) = (t, y(t)), where y(t) 



t sin - if t 0 



0 



iff = 0 



Such a path is called nonrectifiable. It is a fact that any C 1 path with endpoints 
is rectifiable, which is why we made such a condition part of Definition 2.1. 



The Arclength Parameter 

The calculation of the length of a path is not only useful (and moderately inter- 
esting) in itself, but it also provides a way for us to reparametrize the path with 
a parameter that depends solely on the geometry of the curve traced by the path, 
not on the way in which the curve is traced. 

Let x be any C 1 path and assume that the velocity x' is never zero. Fix a point 
P 0 on the path and let a be such that x{a) = Pq. We define a one-variable function 
s of the given parameter t that measures the length of the path from Pq to any 
other (moving) point P by 




Figure 3.19 The arclength 
reparametrization. 




(See Figure 3.19. The Greek letter tau, t, is used purely as a dummy variable — 
the standard convention is never to have the same variable appearing in both the 
integrand and either of the limits of integration.) If t happens to be less than a, 
then the value of s in formula (4) will be negative. This is nothing more than a 
consequence of how the "base point" Po is chosen. 

Here's how to get the new parameter: From formula (4) and from the funda- 
mental theorem of calculus, 



ds 
dt 



d 
dt 



J a 



x\x)\\dx = \\x'(t)\\ = speed. 



(5) 



Since we have assumed that x'(t) ^ 0, it follows that ds/dt is nonzero. Hence, 
ds/dt is always positive, so s is a strictly increasing function of t. Thus, s is, 
in fact, an invertible function; that is, it is at least theoretically possible to solve 
the equation s = s(t) for t in terms of s. If we imagine doing this, then we can 
reparametrize the path x, using the arclength parameter s as independent variable. 

EXAMPLE 3 For the helix x(t ) = (a cost, a sin t, bt), if we choose the "base 
point" P 0 to be x(0) = (a, 0, 0), then we have 

s(t) = f \\x\r)\\dr= f ja 2 + b 2 dx = ja 2 + b 2 t, 
Jo Jo 

so that 

s = y/a 2 + b 2 t, 



Chapter 3 | Vector-Valued Functions 



or 

s 

Va 2 + b 2 ' 

(What the preceding tells us is that this reparametrization just rescales the time 
variable.) Hence, we can rewrite the helical path as 

/ / s \ ( s \ bs 

x(s) = la cos . , a sin . , . 

V \s/a 2 + b 2 ) \y/a 2 + b 2 ) Va 2 + b 2 

EXAMPLE 4 The explicit determination of the arclength parameter for a given 
parametrized path is a delicate matter. Consider the path 

Thenx'(0 = (1, V2t, t 2 ) and, if we take the base point to be x(0) = (0, 0, 0),then 

s(t)= f y/l + 2x 2 + x 4 dx 
Jo 

f' f' ? 3 

= / v / (1 + t 2 ) 2 Jt = / (l + x 2 )dx = t + -. 
Jo Jo 3 

On the other hand the path y(t) = (?, f 2 , f 3 ) is quite similar to x, yet it has 
no readily calculable arclength parameter. In this case, y'(0 = (1, 2f, 3? 2 ) and the 
resulting integral for s(t) is 



s(t)= f y/\+Ax 2 + 9x A dx. 
Jo 



It can be shown that this integral has no "closed form" formula (i.e., a formula 
that involves only finitely many algebraic and transcendental functions). ♦ 

The significance of the arclength parameter s is that it is an intrinsic param- 
eter; it depends only on how the curve itself bends, not on how fast (or slowly) 
the curve is traced. To see more precisely what this means, we resort to the chain 
rule. Consider s as an intermediate variable and t as a final variable. Then we 
have 

ds 

x(t) = x (s) — by the chain rule, 

dt 

= As)\\x'(t)\\ by (5). 
Since x'(t) ^ 0, we can solve for x'(s) to find 

x'(s)= — —. (6) 
l|x'(OII 

Therefore, x'(s) is precisely the normalization of the original velocity vector, and 
so it is a unit vector. Hence, the reparametrized path x(s ) has unit speed, regardless 
of the speed of the original path x(f). (This result makes good geometric sense, 
too. If arclength, rather than time, is the parameter, then speed is measured in 
units of "length per length," which necessarily must be one.) 

The only unfortunate note to our story is that the integral in formula (4) is 
usually impossible to compute exactly, thus making it impossible to compute s 
as a simple function of t. (The case of the helix is a convenient and rather special 



3.2 | Arclength and Differential Geometry 



exception.) One generally prefers to work indirectly, letting the chain rule come 
to the rescue. We shall see this indirect approach next. 



The Unit Tangent Vector and Curvature 

Let x: / c R — ■>• R 3 be a C 3 path and assume that x' is never zero. 



DEFINITION 2.2 The unit tangent vector T of the path x is the normal- 
ization of the velocity vector; that is, 

v x'(0 



T = 



Ml l|x'(OII 



We see from Definition 2.2 that the unit tangent vector is undefined when the 
speed of the path is zero. Also note that, from equation (6), T is dx/ds, where s 
is the arclength parameter. Geometrically, T is the tangent vector of unit length 
that points in the direction of increasing arclength, as suggested by Figure 3.20. 

EXAMPLE 5 For the helix x(f) = (a cost, a sin t, bt), we have 

x'(f) — a sin? i + a cos t j + bk. 

T(f) = — — = J . 

I|x'(0ll Ja 2 + b 2 

On the other hand if we parametrize the helix using arclength so that 

bs 



then 



x(s) = a cos , ] , a sin . 

\Vfl 2 + W \y/a 2 + b 2 ) Va^Tb 2 



T(s) = x'O) = . = = sin i + cos 



■s/a 2 + b 2 V V« 2 + b 2 ) y/a 2 + b 2 \-Ja 2 + b 2 
b 

+ 



Va 2 + b 2 

This agrees (as it should) with the first expression for T, since s = -J a 2 + b 2 t, 
as shown in Example 3. ♦ 

Using the unit tangent vector, we can define a quantity that measures how 
much a path bends as we travel along it. To do so, note the following key facts: 

PROPOSITION 2.3 Assume that the path x always has nonzero speed. Then 

1. dT/dt is perpendicular to T for all t in / (the domain of the path x). 

2. \\dT/dt || \ t=to equals the angular rate of change (as t increases) of the direc- 
tion of T when t = to. 



PROOF (You can omit reading this proof for the moment if you are interested in 
the main flow of ideas.) To prove part 1 , we have 



T(0-T(0 = 1, 



Chapter 3 | Vector-Valued Functions 



since T is a unit vector. Hence, 



T(r 0 + At) 




T(t 0 ) 



Figure 3.21 The vector triangle 
used in the proof of 
Proposition 2.3. 



dt 



(T • T) = 0, 



because the derivative of a constant is zero. Also we have 

d dT dT 
— (T-T) = T 1 T, 

dt dt dt 

by the product rule (Proposition 1.4). Thus, 

dT 

2T = 0. 

dt 

Therefore, T is always perpendicular to dT/dt. (See Proposition 1.7.) 

Now we prove part 2. Because T is a unit vector for all t, only its direction 
can change as t increases. This angular rate of change of T is precisely 

lim — , 

Af->0+ At 

where AO comes from the vector triangle shown in Figure 3.21. To make the 
argument technically simpler, we shall assume that AT ^ 0. We claim that 

AO 



lim 

Ar^0+ IIATI 



1. 



(7) 



Then, from equation (7), 



AO AO ||AT|| 
lim — = lim 

AI-S-0+ At Af^0+ II AT 



At 



= lim 



AO 



lim 



IATII 



a^o+ II AT II Af^o+ At 



= 1 • lim 



IATII 



Ar^0+ At 

Since At is assumed to be positive in the limit, we may conclude that 
lim 





AT 




dT 


— - = lim 






dt 


At Ar^0+ 


At 





as desired. 

To establish equation (7), the law of cosines applied to the vector triangle in 
Figure 3.21 implies 

|| AT|| 2 = ||T(f + Af)|| 2 + ||T(f)|| 2 - 2||T(t + At)|| ||T(t)|| cos AO 

= 2 - 2 cos AO, 

because T is always a unit vector. Thus, 

AO 



lim 



lim 



AO 



Ar^o+ ||AT|| a^o+ 72 - 2 cos AG 

AO 

= lim - 

A '^ 0+ ^2 • 2(sin 2 (A6>/2)) 

from the half-angle formula, and so 

AO AO/2 

lim = lim = 1, 

Ar^o+ ||AT|| Ar^o+ sin(A<9/2) 

from the well-known trigonometric limit (or from L'Hopital's rule). 



3.2 j Arclength and Differential Geometry 209 

Part 2 of Proposition 2.3 provides a precise way of measuring the bending of 
a path. 



DEFINITION 2.4 The curvature k of a path x in R 3 is the angular rate of 
change of the direction of T per unit change in distance along the path. 



The reason for taking the rate of change of T per unit change in distance in the 
definition of k is so that the curvature is an intrinsic quantity (which we certainly 
want it to be). Figure 3.22 should help you develop some intuition about k. 




Figure 3.22 In the left figure, k is not large, since the 
path's unit tangent vector turns only a small amount per 
unit change in distance along the path. In the right 
figure, k is much larger, because T turns a great deal 
relative to distance traveled. 



Because ||dT/<2r|| measures the angular rate of change of the direction of T 
per unit change in parameter (by part 2 of Proposition 2.3) and ds/dt is the rate 
of change of distance per unit change in parameter, we see that 




where the last equality holds by the chain rule. It is formula (8) that we will use 
when making calculations. 

EXAMPLE 6 For the circle x(?) = {a cos t, a sin t), 0 < t <2tt, 

ds 

x (?) = — a sin? i + a cost j, ||x (?)|| = — = a, 

dt 

so that 

xYf) 

T(?) = = — sin? i + cos? j. 

I|x'(?)|| 



Hence, 



\\dT/dt\\ 1 . 1 

k = = -|| -cos?i-sm?j|| = - 

ds/dt a a 



Thus, we see that the curvature of a circle is always constant with value equal 
to the reciprocal of the radius. Therefore, the smaller the circle, the greater the 
curvature. (Draw a sketch to convince yourself.) ♦ 



210 



Chapter 3 | Vector-Valued Functions 



EXAMPLE 7 If a and b are constant vectors in R 3 and the path 

x(t) = a t + b 



traces a line. We have 



so 



x'(0 = a, 
ds 



Hence, 



dt 



T(/) = 



= a . 



which is a constant vector. Thus, T'(f) = 0 and formula (8) implies immedi- 
ately that k is zero, which agrees with the intuitive fact that a line doesn't 
curve. ♦ 

EXAMPLE 8 Returning to our friend the helix 

x(t) = (a cos t , a sin t , bt), 

we have already seen that 



ds 
dt 



= ja 2 + b 2 and T(f) = 



-a sin / i + a cos / j + b k 

Va 2 + b 2 



Thus, formula (8) gives 



1 



sja 2 + b 2 



-a cos / i — a sin? j 



Va 2 + b 2 



a 2 + b 2 ' 



We see that the curvature of the helix is constant, just like the circle. In fact, as b 
approaches zero, the helix degenerates to a circle, and the resulting curvature is 
consistent with that of Example 6. 

We can also compute the curvature from the parametrization given by arc- 
length. The same helix is also described by 



\(s) = a cos 



Vfl 2 + b 2 



, a sin 



bs 



vV + W Va 2 + W 



and we have 
dx 



T(s) = 



ds 



sin 



+ 



Va 2 + b 2 \Ja 2 + b 2 
b 



+ 



cos 



Vfl 2 + b 2 \y/a 2 + b 2 



■s/a 2 + b 2 
We can, therefore, compute 
dT a 
ds 



cos 



a 2 + b 2 \Ja 2 + b 2 
and hence, from formula (8), that 

dT 



sin 



+ b 2 \s/a 2 + b 2 



h 



ds 



a 2 + b 2 



which checks. 



3.2 | Arclength and Differential Geometry 



The Moving Frame and Torsion 

We now introduce a triple of mutually orthogonal unit vectors that "travel" with a 
given path x: / —> R 3 , known as the moving frame of the path. (Note: In general, 
the term "frame" means an ordered collection of mutually orthogonal unit vectors 
in R".) These vectors should be thought of as a set of special vector "coordinate 
axes" that move from point to point along the path. 

To begin, assume that (i) x'(0 # 0 and (ii) x'(t) x x"(f) ^ 0 for all t in /. 
(The first condition assures us that x never has zero speed and the second that x 
is not a straight-line path.) Then the first vector of the moving frame is just the 
unit tangent vector: 

dx = x'(Q 
ds ||x'(OII ' 

(Now you see why condition (i) is needed.) For a second vector orthogonal to T, 
recall that part 1 of Proposition 2.3 says that dT/dt must be perpendicular to T. 
Hence, we define 




(That dl/dt is not zero follows from assumptions (i) and (ii).) The vector N is 
called the principal normal vector of x. By the chain rule, N is also given by 

dT/ds 

N = . (10) 

\\dT/ds\\ 

Since k = \\dT/ds\\ by formula (8), we also see that 




At a given point P along the path, the vectors T and N (and also the vectors 
x' and x") determine what is called the osculating plane of the path at P. (See 
Figure 3.23.) This is the plane that "instantaneously" contains the path at P. (More 




Osculating plane 



Figure 3.23 The osculating plane of the path x at the 
point P. 



212 Chapter 3 | Vector- Valued Functions 



precisely, it is the plane obtained by taking points P\ and P2 on the path near P 
and finding the limiting position of the plane through P, Pi, and P% as Pi and 
P2 approach P along x. The word "osculating" derives from the Latin osculare, 
meaning "to kiss.") 

Now that we have defined two orthogonal unit vectors T and N, we can 
produce a third unit vector perpendicular to both: 



B = T x N. 



(12) 



The vector B, called the binormal vector, is defined so that the ordered triple 
(T, N, B) is a right-handed system. Thus, B is a unit vector since 



IBM = II T II 



sin - = 1 ■ 1 • 1 = 1. 

2 



EXAMPLE 9 For the helix x(/) = (a cos t, a sin t, bt), the moving frame vec- 
tors are 



T(f) = 



-a sin t i + a cos t j + b k 

■J a 1 + b 2 



(as we have already seen), 



T'(f) (—a cos t i — a sinf j)/V« 2 + b 2 
fs(t) = — — = ^=^= = — cos 1 1 — sin t j , 



|T'(t)ll 



a/Va 2 + b 2 



and 



B(0 = T x N = 



i j k 

-a sinf/Va 2 + b z a cos?/V« 2 + b 2 b/-Ja 2 + b 2 
— cos t — sin t 0 



sinf i- , , Fi cosf j+ / i k - 



\«Ja 2 + b 2 J \V<3 2 + b 2 J Wfl 2 + b 2 

Equation (11) says that the derivative of T (with respect to arclength) is a 
scalar function (namely, the curvature) multiple of the principal normal N. This 
is not surprising, since N is defined to be parallel to the derivative of T. A more 
remarkable result (see the addendum at the end of this section) is that the derivative 
of the binormal vector is also always parallel to the principal normal; that is, 

dB 

— = (scalar function) N. 

ds 

The standard convention is to write this scalar function with a negative sign, so 
we have 



dB 

— = -tN. (13) 
ds 



The scalar function r thus defined is called the torsion of the path x. Roughly 
speaking, the torsion measures how much the path twists out of the plane, how 



3.2 | Arclength and Differential Geometry 



"three-dimensional" x is. Note that, according to our conventions, the curvature 
k is always nonnegative (why?), while r can be positive, negative, or zero. 

EXAMPLE 10 Consider again the case of circular motion. Thus, let x(/) = 
(a cost, a sin /). Then, as shown in Example 6, 



T(0 



x'(t) 



|x'(OII 



sin? i + cos? j, and k 



dT 



ds 



Now we calculate 



T'(0 

N = = — cos M-sinn', 

l|T'(0ll 

B = TxN = k, a constant vector. 

Hence, dB/ds = 0, so there is no torsion. This makes sense, since a circle does 
not twist out of the plane. ♦ 

EXAMPLE 1 1 Let x(/) = (e' cos /, e' sin / , e'). We calculate T, N, and B and 
identify the curvature and torsion of x. 
To begin, we have 

x'(t) e'(cos/ — sin/)i + e'(cos/ + sin/)j + e' k 

||x'(0|| = Vfe< 



1 

71 



((cos t — sin t) i + (cos t + sin t ) j + k) . 



From this, we may compute 

dT dT/dt ^(-(sin? + cos0i + (cosf - sinf)j) 
ds ds/dt s/3e' 

e-' 

= — (— (sinf + cosf)i + (cos t — sinf)j), 



so that the curvature is 



dT 



ds 



sflt 



3 



Now we determine the remainder of the moving frame: 
T'(0 1 

N = -t: jr = —=(— (sinf + cos t)i + (cosf — sin/) j), 

T'(0 V2 



B = T x N = —((sinf - cos/)i - (sin/ + cos/)j + 2k). 
V 6 

Finally, to find the torsion, we calculate 

dB _ dB/dt _ ^((cos/ + sin/)i + (sin/ -cos/)j) 
ds ds/dt yfZe* 

((cos f + sin /) i + (sin / — cos /) j) 



3^2 



-N, 



so 



T = 



214 Chapter 3 | Vector-Valued Functions 




Figure 3.24 Any vector in the 
plane perpendicular to T can be 
used for N. 



EXAMPLE 12 If a and b are vectors in R 3 , then the straight-line path x(t) = 
at + bhas, as we saw in Example 7, T = a/||a||. Thus, both dT/df anddT/ds are 
identically zero. Hence, k = 0 (as shown in Example 7) and N cannot be defined 
using formula (9). From geometric considerations, any unit vector perpendicular 
to T can, in principle, be used for N. (See Figure 3.24.) If we choose one such 
vector, then B can be calculated from formula (12). Since T, N, and B are all 
constant, r must be zero. This is an example of a moving frame that is not 
uniquely determined by the path x and serves to illustrate why the assumption 
x' x x" / 0 was made. ♦ 

It is important to realize that the moving frame, curvature, and torsion are 
quantities that are intrinsic to the curve traced by the path. That is, any parame- 
trized path that traces the same curve (in the same direction) must necessarily 
have the same T, N, B vector functions and the same curvature and torsion. This 
is because all of these quantities can be denned entirely in terms of the intrinsic 
arclength parameter s. (See Definition 2.2 and formulas (6), (8), (10), (1 1), (12), 
and (13).) 

Another important fact is that the curvature function k and the torsion function 
t together determine all the geometric information regarding the shape of the 
curve, except for the curve's particular position in space. To be more precise, we 
have the following theorem, whose proof we omit: 



THEOREM 2.5 Let s be the arclength parameter and suppose C\ and C2 are 
two curves of class C 3 in R 3 . Assume that the corresponding curvature functions 
k\ and K2 are strictly positive. Then if k\(s) = K2(s) and t\(s) = t2(s), the two 
curves must be congruent (in the sense of high school geometry). In fact, given 
any two continuous functions k and r, where k{s) > 0 for all s in the closed 
interval [0, L], there is a unique curve parametrized by arclength on [0, L] (up to 
position in space) whose curvature and torsion are k and r, respectively. 



Tangential and Normal Components of Velocity and Accel- 
eration; Other Curvature Formulas — 

As we have seen, the moving frame provides us with an intrinsic set of vectors, 
like coordinate axes, that are special to the particular curve traced by a path. In 
contrast, the velocity and acceleration vectors of a path are definitely not intrinsic 
quantities but depend on the particular parametrization chosen as well as on the 
shape of the path. (The speed of a path is entirely independent of the geometry 
of the curve traced.) We can get some feeling for the relationship between the 
intrinsic notion of the moving frame and the extrinsic quantities of velocity and 
acceleration by expressing the latter two vector functions in terms of the moving 
frame vectors. 

Thus, we begin with a C 2 path x: / — > R 3 having x' 7^ 0 and x' x x" ^ 0. 
For notational convenience, let s denote ds/dt and s denote d 2 s/dt 2 . From 
Definition 2.2, we know that T = v/||v|| and so, since the speed s = ds/dt = ||v||, 
we have 



v(f) = sT. (14) 



3.2 | Arclength and Differential Geometry 




Figure 3.25 Decomposition of 
acceleration a into tangential and 
normal components. 



This formula says that the velocity is always parallel to the unit tangent vector, 
something we know well. To obtain a similar result for acceleration, we can 
differentiate (14) and apply the product rule: 



a(0 = V(t) 



dt 



(sT) = ST + s 



dT 
dt ' 



(15) 



Next, we express dT/dt in terms of the T, N, B frame. Formula (11) gives 
the derivative of dT/ds in terms of N. The chain rule says that dT/ds = 
(dT/dt)/(ds/dt). Thus, from formula (1 1), we have 

dT dT 

—— = s— = skN. 

dt ds 

Hence, we may rewrite equation (15) as 



a(r) = j'T + k j 2 N. 



(16) 



Warning s = d 2 s/ dt 2 is the derivative of the speed which is a scalar function. 
The acceleration a is the derivative of velocity and so is a vector function. 

Note that formula (16) shows that the acceleration has no component in the 
direction of the binormal vector B. Therefore, both velocity and acceleration are 
vectors that lie in the osculating plane of the path. (See Figure 3.25.) 

At first glance, it may not appear to be especially easy to use formula (16) 
to resolve acceleration into its tangential and normal components because of the 
curvature term. However, 

||a|| 2 = a • a = (j'T + k j 2 N) • (ST + /fj 2 N) = s 2 + (ks 2 ) 2 , 

since T and N are perpendicular vectors. Consequently, we may calculate the 
components as follows: 



Tangential component of acceleration = a tang = s. 



Normal component of acceleration = a n0 rm = ks 2 = J ||a|| z — a 



'tang- 



EXAMPLE 1 3 Let x( Q = (t , I t , t 2 ). Then v(/) = i + 2j + 2/k and a(f ) = 2k. 

We have s = ||v(f)ll = V5 + 4f 2 . Therefore, 



At 



V5 + At 2 ' 
Since ||a|| = 2, we see that 



I a II 2 - fitting 



16/ 2 2^5 



5 + 4f 2 V5 + At 2 ' 



Formulas (14) and (16) enable us to find an alternative equation for the 
curvature of the path. We simply calculate that 

v x a = (iT) x (sT + /ci 2 N) = ss(T x T) + kP(T x N) = ks 3 B. 



216 Chapter 3 I Vector- Valued Functions 

Recalling that s = ||v||, we have, by taking magnitudes, 
||v x a|| = k||v|| 3 ||B|| = K ||v|| 3 , 
since B is a unit vector. Thus, 



||vxa|| 


(17) 


K = z—. 

IMP 





This relatively simple formula expresses the curvature (an intrinsic quantity) 
in terms of the nonintrinsic quantities of velocity and acceleration. 

EXAMPLE 14 For the path \(t) = (2t 3 + l,t 4 , t 5 ), we have 



and 



You can check that 



v(0 = 6th + 4f 3 j + 5r 4 k 



a(0 = 12ri + I2t 2 \ + 20f 3 k. 



|v|| = t 2 ^25t 4 + I6t 2 + 36 



and 



||v x a|| = ||4f 4 (5f 2 i - I5tj + 6k)|| = 4fV25f 4 + 225/ 2 + 36. 

Therefore, formula (17) yields 

|| v x a|| _ 4(25f 4 + 225r 2 + 36) 1/2 
* ~ ||v|| 3 ~ t 2 (25t 4 + 16f 2 + 36) 3 / 2 ' 

which is certainly a more convenient way to determine curvature in this case. ♦ 



Summary 

You have seen many formulas in this section, and, at first, it may seem difficult 
to sort out the primary statements from the secondary results. We list the more 
fundamental facts here: 

For a path x: / — >• R 3 : 



Nonintrinsic quantities: 

Velocity v(0 = x'(t). 

Speed % = ||v(/)||. 
dt 

Acceleration a(t) = x"(?). 



3.2 | Arclength and Differential Geometry 



Arclength function: (See Figure 3.26.) 

s(t) = / ||x'(t)||Jt (basepoint is Pq = x(a)) 

J a 



Intrinsic quantities: 

The moving frame: 



Unit tangent vector T = 

Principal normal vector N = 
Binormal vector B = 

Curvature k = 
dB 



Torsion r is denned so that 



ds 



dx 


X 


ds~ ~ 


II* 


dT/ds 


\\dT/ds\\ 


TxN. 


dT 




ds 




-tN 





At) 



dT/dt 
= \\dT/dt\ 

WdT/dtW 
ds/dt 



Additional formulas: 

v(?) = s T (s is speed). 

a(t) = ST + ks 2 N (s is derivative of speed). 
Ilv x all 



Addendum: More About Torsion and the 
Frenet-Serret Formulas 

We now derive formula (13), the basis for the definition of the torsion of a curve. 
That is, we show that the derivative of the binormal vector B (with respect to 
arclength) is always parallel to the principal normal N (i.e., that dB/ds is a 
scalar function times N). The two main ingredients in our derivation are part 1 of 
Proposition 2.3 and the product rule. 

We begin by noting that, since the ordered triple of vectors (T, N, B) forms a 
frame for R 3 , any moving vector, including dB/ds, can be expressed as a linear 
combination of these vectors; that is, we must have 



dB 

ds 



a(s)T + Z?(s)N + c(«)B, 



(18) 



where a, b, and c are appropriate scalar- valued functions. (Because T, N, and 
B are mutually perpendicular unit vectors, any (moving) vector w in R 3 can be 
decomposed into its components with respect to T, N, and B in much the same 
way that it can be decomposed into i, j, and k components — see Figure 3.27.) To 
find the particular values of the component functions a, b, and c, it turns out that 



218 Chapter 3 | Vector-Valued Functions 



we can solve for each function by applying appropriate dot products to equation 
(18). Specifically, 

dB 

T = a(s)T • T + b(s)N • T + c(s)B ■ T 

ds 

= a(s) ■ 1 + b(s) • 0 + c(s) • 0 
= a(s), 

and, similarly, 

dB dB 

— -N = —.B = c(s). 

ds as 

From Proposition 1.7, dB/ds is perpendicular to B and, hence, c must be zero. 
To find a, we use an ingenious trick with the product rule: Because T • B = 0, it 
follows that d /ds(T • B) = 0. Now, by the product rule, 

d dB dT 
— (TB) = T 1 B. 

ds ds ds 

Consequently, (dB/ds) • T = —(dT/ds) • B. Thus, 

, . dB ^ dT „ 

a(s) = T = B 

ds ds 

= -*rN-B by formula (11), 
= 0, 



and equation ( 1 8) reduces to 



dB 

= b(sys. 

ds 



No further reductions are possible, and we have proved that the derivative of B is 
parallel to N. The torsion r can, therefore, be defined by r(s) = — b(s). 

Formulas (1 1) and (13) gave us intrinsic expressions for dT/ds and dB/ds, 
respectively. We can complete the set by finding an expression for dN/ds. The 
method is the same as the one just used. Begin by writing 

JN 

— = o(j)T + MsyS + c(s)B, (19) 
ds 

where a, b, and c are suitable scalar functions. Taking the dot product of equation 
(19) with, in turn, T, N, and B, yields the following: 

JN 

a(s) = — • T, b(s) = — • N, c(s) = — • B. 
ds ds ds 

The "product rule trick" used here then reveals that 

i/N „ dT 

a(s)= — .T = -N.— 
ds ds 

= — N-/cN by formula (11) 



and 



JN JB 
c(s)= B = -N.— 
ds 



= — N • (-tN) by formula (13) 

= T. 



3.2 | Exercises 219 



Moreover, we may differentiate the equation N • N = 1 to find 

b(s) 



dN dN 
■ N = — N • 



= -a:T + tB. 



ds ds ' 

which implies that b(s) is zero. Hence, equation (19) becomes 

dN 

ds 

The formulas for dT/ds, dN/ds, and dB/ds are usually taken together as 

T'O) = kN 
N'(i) = -kT + tB 
B'O) = -tN 

and are known as the Frenet-Serret formulas for a curve in space. They are so 
named for Frederic-Jean Frenet and Joseph Alfred Serret, who published them 
separately in 1852 and 1851, respectively The Frenet-Serret formulas give a 
system of differential equations for a curve and are key to proving a result like 
Theorem 2.5. They are often written in matrix form, in which case, they have an 
especially appealing appearance, namely, 



r " 




0 


K 


0 " 




~ T 


N' 




— K 


0 


X 




N 


B' 




0 


— T 


0 




B 



3.2 Exercises 



Calculate the length of each of the paths given in Exercises 
1-6. 



1. 


x(f) = 


(2t+ l,7-3*),-l <r <2 


2. 


x(f) = 


f 2 i+ |(2f + l) 3 / 2 j, 0 < f < 4 


3. 


x(f) = 


(cos 3?, sin3f, 2? 3 / 2 ), 0 < t < 2 


4. 


x(f) = 


7i + t] + t 2 k, 1 < t < 3 


5. 


x(f) = 


(r 3 ,3r 2 ,6f), -1 <r <2 


6. 


x(f) = 


(In (cos 0, cost, sin f), f < ? < f 


7. 


x(f) = 


Qnt, t 2 /2, y/2t), 1 < t < 4 


8. 


x(f) = 


(2t cosf,2f sinr,2V2f 2 ), 0 < ? < 3 


9. 


The path x(f ) = (a cos 3 1, a sin 3 f), where a 



tive constant, traces a curve known as an astroid or a 
hypocycloid of four cusps. Sketch this curve and find 
its total length. (Be careful when you do this.) 

10. If / is a continuously differentiable function, show 
how Definition 2.1 may be used to establish the 
formula 

L = / y/\+{f'( X )Ydx 
J a 



forthe length ofthe curve y = f(x) between (a, f(a)) 
and (b, f(b)). 

1 1 . Use Exercise 10 or Definition 2.1 (or both) to calculate 
the length of the line segment y = mx + b between 
(xq, yo) and (xi, yi). Explain your result with an ap- 
propriate sketch. 

12. (a) Calculate the length of the line segment deter- 

mined by the path 

x(t) = (a l t + b u a 2 t + b 2 ) 

as f varies from to to t\ . 

(b) Compare your result with that of Exercise 1 1 . 

(c) Now calculate the length of the line segment deter- 
mined by the path x(f ) = a t + b as t varies from 
f 0 to h . 

13. This problem concerns the path x=|f — l|i+|r|j, 

-2 < t < 2. 

(a) Sketch this path. 

(b) The path fails to be of class C 1 but is piecewise 
C l . Explain. 

(c) Calculate the length of the path. 

14. Consider the path x(f) = (e~' cost, e~' sinf). 



Chapter 3 | Vector-Valued Functions 



(a) Argue that the path spirals toward the origin as 
t -»- +00. 

(b) Show that, for any a, the improper integral 



r 

J a 



\W{t)\\dt 



converges. 

(c) Interpret what the result in part (b) says about the 
path x. 

1 5. Suppose that a curve is given in polar coordinates by 
an equation of the form r = f(9), where / is of class 
C 1 . Use Definition 2.1 to derive the formula 



J a 



(8) 2 + f(6) 2 d6 



for the length of the curve between the points (/(a), a) 
and (/(/S), j3) (given in polar coordinates). 

16. (a) Find the arclength parameter s = s(t) for the path 

x(?) = e at cos bt i + e at sin bt j + e at k. 

(b) Express the original parameter / in terms of s and, 
thereby, reparametrize x in terms of s. 

Determine the moving frame {T, N, B}, and compute the cur- 
vature and torsion for the paths given in Exercises 1 7—20. 

17. x(0 = 5cos3H + 6t j + 5 sin 3? k 

18. x(f) = (sin? - t cos?)i + (cosf + t sinf)j + 2k, 
t > 0 

19. x(r) = (t, \(t + I) 3 / 2 , 1(1 - tf' 2 ), -1 < t < 1 

20. x(f) = (e 2 ' sin/, e 2 ' cosf, 1) 

21. (a) Use formula (17) in this section to establish the 

following well-known formula for the curvature 
of a plane curve y = f(x): 



[i + (/'(x)) 2 F 2 ' 

(Assume that / is of class C 2 .) 

(b) Use your result in (a) to find the curvature of 
y = In (shut). 

22. (a) Let x(s) = {x(s), y(s)) be a plane curve para- 
metrized by arclength. Show that the curvature is 
given by the formula 

K = \x'y" -x"y'\. 



path x and, separately, plot the curvature k as a function of t 
over the indicated interval for t and value(s) of the constants. 

^ 23. x(?) = (acos?, fcsinr), 0 < t < 2jt; a = 2,b=l 

^24. x(r) = (2a(l + cos r) cos f, 2a(l + cos f) sin f), 0< 

t < 2jt; a = 1 

^ 25. x(r) = (2a cosf(l + cost) - a, 2a sinf(l + cosf)), 
0 < t < 2ji; a = l 

^^26. x(r) = (a sin nt, b sin mt), 0 < t < 2jt; a = 3, 
b = 2, n = 4, m = 3 

Find the tangential and normal components of acceleration for 
the paths given in Exercises 27—32. 



(b) Showthatx(s)=(i(l-.y 2 ), ^cos" 1 s - Wl - s 2 

is parametrized by arclength, and compute its 
curvature. 

In Exercises 23-26, (a) use a computer algebra system to cal- 
culate the curvature k of the indicated path x and (b) plot the 



>) 



27. x(?) = 


f 2 i + fj 


28. x(?) = 


(2t , e 2 ') 


29. x(r) = 


(e' cos2f , e* sin 2?) 


30. x(r) = 


(4cos5f, 5 sin4f, 30 


31. x(?) = 


(t,t,r 2 ) 


32. x(?) = 


1(1 — cos t )i + sinf j + ^ cos t k 



33. (a) Show that the tangential and normal compo- 
nents of acceleration a tang and a n orm satisfy the 
equations 



x x x 



^tang 



(b) Use these formulas to find the tangential and 
normal components of acceleration for the path 
x(f) = (?+2)i + f 2 j + 3fk 

34. Use Exercise 33 to show that, for the plane curve 

y = m, 

f'(x)f"(x) 
"tang — , ) 

yi+(/'«) 2 

\f"{*)\ 
^norm — ; ■ 

yi + (f'M) 2 

35. Establish the following formula for the torsion: 

(v x a) • a' 
II v x a[| 2 

36. Show that kx = — T' • B', where differentiation is with 
respect to the arclength parameter s. 

37. Show that if x is a path parametrized by arclength and 
x' x x" ^ 0, then 

K 2 T = (X'XX")-X"'. 

38. Suppose x: / -> R 3 is a path with x'(f) x x"(r) / Ofor 
all t e /. The osculating plane to the path at t = to is 
the plane containing x(/o) and determined by (i.e., par- 
allel to) the tangent and normal vectors T(fo) and N(fo). 



3.3 | Vector Fields: An Introduction 



The rectifying plane at t = to is the plane contain- 
ing x(fo) and determined by the tangent and binormal 
vectors T(fo) and B(?o). Finally, the normal plane at 
t = to is the plane containing x(fo) and determined by 
the normal and binormal vectors N(?o) and B(fo). Note 
that both the osculating and rectifying planes may be 
considered to be tangent planes to the path at to since 
they are both parallel to T(fo). 

(a) Show that B(fo) is perpendicular to the osculating 
plane at to, that N(?o) is perpendicular to the rec- 
tifying plane at to, and that T(to) is perpendicular 
to the normal plane at to- 

(b) Calculate the equations for the osculating, rec- 
tifying, and normal planes to the helix x(?) = 
(a cos t, a sin?, bi) at any ?o- (Hint: To speed your 
calculations, use the results of Example 9.) 

39. Recall that the equation for a sphere of radius a > 0 
and center xo may be written as ||x — xo[| = a. (See 
Example 15 of §2. 1 .) Explain why the image of a path 
x with the property that 

(x(f) - x 0 ) • (x(r) - x 0 ) = a 2 

for all t must lie on a sphere of radius a. 

40. Let x be a path with x' x j" / 0 and suppose that there 
is a point xo that lies on every normal plane to x. Show 
that the image of x lies on a sphere. (See Exercise 38 
concerning normal planes to paths.) 

41 . Use the result of Exercise 40 to show that x(? ) = 
(cos2f, — sin2f, 2 cos/) lies on a sphere by showing 
that (1,0,0) lies on every normal plane to x. 

42. Use the result of Exercise 27 of § 1 .4 to show that 

N x B = T and B x T = N. 



As a result, we can arrange T, N, and B in a circle so 
that they correspond, respectively, to the vectors i, j, 
k appearing in Figure 1.54 and so that we may use a 
mnemonic for identifying cross products that is similar 
to the one described in Example 1 of § 1 .4. 

Let x be a path of class C 3 , parametrized by arclength s, 
with x' x x" / 0. We define the Darboux rotation vector (also 
called the angular velocity vector) by 

w = rT + x:B. 

Note that w(.?o) is parallel to the rectifying plane to x(so). The 
direction of the Darboux vector w gives the axis of the "screw- 
like" motion of the path x and its length gives the angular 
velocity of the motion. Exercises 43-45 concern the Darboux 
vector. 

43. Show that ||w[| = si k 1 + r 2 . (Hint: The vectors T, N, 
and B are pairwise orthogonal.) 

44. (a) Use the Frenet-Serret formulas to establish the 

Darboux formulas: 

T' = w x T 
N' = w x N 
B' = w x B. 

(b) Use the Darboux formulas to establish the Frenet- 
Serret formulas. Hence the two sets of equations 
are equivalent. (Hint: Use Exercise 42.) 

45. Show that x is a helix if and only if w is a constant 
vector. (Hint: Consider w' and use Theorem 2.5.) 



/ / / / 



Figure 3.28 The constant vector 
field F(x) = i + j. 



3.3 Vector Fields: An Introduction 

We begin with a simple definition. 



DEFINITION 3.1 A vector field on R" is a mapping 

MC R" R". 



We are concerned primarily with vector fields on R 2 or R 3 . fn such cases, we 
adopt the point of view that a vector field assigns to each point x in X a vector 
F(x) in R" , represented by an arrow whose tail is at the point x. This perspective 
allows us to visualize vector fields in a reasonable way. 

EXAMPLE 1 Suppose F: R 2 -■>• R 2 is defined by F(x) = a, where a is a con- 
stant vector. Then F assigns a to each point of R 2 , and so we can picture F by 
drawing the same vector (parallel translated of course) emanating from each point 
in the plane, as suggested by Figure 3.28. ♦ 



Chapter 3 | Vector-Valued Functions 



(x, y) 


G(x, y) 


(0,0) 


0 


(1,0) 


-j 


(0,1) 


i 


(1,1) 


i- j 




Figure 3.29 The vector field 
G(x, y) = yi — xj of Example 2. 



EXAMPLE 2 Let's depict G: R 2 -> R 2 , G(x, y) = yi- xj. We can begin to do 
this by calculating some specific values of G, as in the adjacent table. However, 
it is difficult to get much of a feeling for G as a whole in this way. To understand 
G somewhat better, we need to "play around" a bit. Note that 



IGCoOII = llvi-xjl 



= r 



where r = xi + yj, the position vector of the point (x, y). From this observation, 
it follows that G has constant length a on the circle x 2 + y 2 = a 2 . In addition, 
we have 



r • G(x, y) = (xi + yj) • (yi - xj) = 0. 

Hence, G(x, y) is always perpendicular to the position vector of the point (x, y). 
These facts, together with a table like the preceding one, make it possible to see 
that G looks like Figure 3.29. ♦ 



Remark Sometimes a scalar- valued function f:X c R" — >• R is called a scalar 
field. One thinks of a vector field on R" as attaching vector information (such 
as wind velocity) to each point and a scalar field as attaching real number infor- 
mation (such as temperature or pressure). We'll use the term "scalar field" only 
occasionally, but we don't want to shock you when we do. 

EXAMPLE 3 Let r = xi + yj + zk. The so-called inverse square vector field 
in R 3 is a function F: R 3 — {0} — > R 3 given by 



F(x, y, z) = 



where c is any (nonzero) constant. If the term "inverse square" seems inappropriate 
to you, we'll try to convince you otherwise. Setu = r/||r|| sothatr = ||r||u. Then 
F is given by 



F(x, y, z) = 




Figure 3.30 An inverse square 
vector field. 



(1) 



Therefore, F is a vector field whose direction at the point P(x, y, z) ^ (0, 0, 0) 
is parallel to the vector from the origin to P and whose magnitude is inversely 
proportional to the square of the distance from the origin to P. Note that F points 
away from the origin if c is positive and toward the origin if c is negative. 

We have seen an example of an inverse square field in §3.1 — namely, the 
Newtonian gravitational field between two bodies. If one of the bodies is at the 
origin and the other at (x, y, z), then we have 

GMm 
F= ^u. 



In this case, the proportionality constant c is —GMm, which is negative. This 
means that the gravitational force is attractive (i.e., it points in the direction 
that reduces the distance between the two bodies). Such a vector field is shown in 
Figure 3 .3 0. An example of a repelling inverse square field is the electrostatic force 



3.3 | Vector Fields: An Introduction 



between two particles with like static charges (both positive or both negative). 
This force is expressed by Coulomb's law, 

F = z- u, 



where r is the vector from particle 1 (at the origin) to particle 2, u = r/||r||, 
q\ and qi are the respective charges (positive or negative) on the particles, and 
k is a constant appropriate for the units being used. In mks units, distance is 
measured in meters, charge in coulombs, force in newtons, so that k is equal to 
8.9875 x 10 9 Nm 2 /C 2 . ♦ 



Gradient Fields and Potentials 

Inverse square fields are interesting not only for their origin in basic physical 
situations, but also because they are examples of gradient fields. A gradient field 
on R" is a vector field F: X c R" — > R" such that F is the gradient of some 
(differentiable) scalar- valued function /: X -> R. That is, 

F(x) = V/(x) 

at all x in X. The function / is called a (scalar) potential function for the vector 
field F. To see what this means in the case of the inverse square field (1), we write 
out the components of F explicitly: 

xi + yj + zk \ 



||r|| 2 V* 2 + y 2 + z 2 J \j x 2 + y 2 + z 2 
since r = xi + yj + zk and u = r/||r||. That is, 

(X ' y ' Z) ~ (x 2 + y 2 + z 2 ) 3 / 2 1 + (x 2 + y 2 + z 2 ) 3 / 2 J + (x 2 + y 2 + z 2 ) 3 / 2 ' 

We leave it to you to check that F(x, y, z) = V/(x, y, z), where /:R 3 — {0} -> R 
is given by 



f(x, y,z) 



c 



y/x 2 + y 2 + z 2 Ilrll 



Remark In physics and engineering, a negative sign is often introduced in the 
definition of a potential function (i.e., so that a potential function g for a vector 
field F is one such that F = — Vg). The motivation behind such a convention is 
that in physical applications, it is desirable to have the potential function rep- 
resent potential energy in some sense. For example, in the case of the gravita- 
tional field F = — ( G M/w / 1 1 r 1 1 2 )u, a physicist would take the potential function to 
be —GMm/\\r\\, not +GMm/||r|| as we do. The advantage to the physicist in 
doing so is that the physicist's potential function increases with increasing ||r||. 
This corresponds to the notion that the greater the distance between two bodies, 
the greater should be the stored gravitational potential energy. 

From Theorem 6.4 of Chapter 2 we know that the gradient of any C 1 scalar- 
valued function /: X c R" ->• R is perpendicular to the level sets of /. Thus, if 
F is a gradient vector field on R" , F(x) must be perpendicular to the level set of a 



224 Chapter 3 I Vector- Valued Functions 




Figure 3.31 A gradient vector field F = V/. Equipotential 
lines are shown where / is constant. 



potential function of F containing the point x. If / is such a potential function, the 
level set {x | f(x) = c] is called an equipotential set (or equipotential surface if 
n = 3, or equipotential line if n = 2) of the vector field F. (See Figure 3.31.) 

You've seen examples of equipotential lines every time you've looked at a 
weather map. Usually curves of constant barometric pressure (called isobars) or 
of constant temperature (isotherms) are drawn. ( See Figure 3.32.) Perpendicular 
to such equipotential lines are associated gradient vector fields that point in the 
direction of most rapid increase of pressure or temperature. 

Flow Lines of Vector Fields 

When you draw a sketch of a vector field on R 2 or R 3 , it is easy to imagine 
that the arrows represent the velocity of some fluid moving through space as in 
Figure 3.33. It's natural to let the arrows blend into complete curves. What you're 




Figure 3.32 A weather map. (Weather graphics courtesy of Accuweather, Inc. 385 
Science Park Road, State College, PA 16803. (814) 237-0309. © 2011. Used with 
permission.) 



Figure 3.33 A fluid moving 
through space. 



3.3 I Vector Fields: An Introduction 



225 




doing analytically is drawing paths whose velocity vectors coincide with those of 
the vector field. 



Figure 3.34 A flow line. 




Figure 3.35 The vector field 
F(x,>>,z) = 2i-3j + kof 
Example 4. 




Figure 3.36 Flow lines of 

F(x, y) = —yi + x\ of Example 5. 



DEFINITION 3.2 A flow line of a vector field F: X C R" -> R" is a differ- 
entiable path x: / — > R" such that 

x'(t) = F(x(0). 

That is, the velocity vector of x at time ? is given by the value of the vector 
field F at the point on x at time ?. (See Figure 3.34.) 



EXAMPLE 4 We calculate the flow lines of the constant vector field 
F(x,y,z) = 2i-3j + k. 

A picture of this vector field (see Figure 3.35) makes it easy to believe that 
the flow lines are straight-line paths. Indeed if x(?) = (x(t), y(t), z(t)) is a flow 
line, then, by Definition 3.2, we must have 

x'(?) = (x'(?), y'(t), z'{t)) = (2, -3, 1) = F(x(?)). 

Equating components, we see 

V(0 = 2 
y'(t) = -3. 
z'(t)= 1 

These differential equations are readily solved by direct integration; we obtain 

x(?) = 2? + xq 
y{t) = -3? + y 0 , 
z(t) = t + zo 

where xq, yo, and zo are arbitrary constants. Hence, as expected we obtain para- 
metric equations for a straight-line path through an arbitrary point (x 0 , yo, Zo) 
with velocity vector (2,-3,1). ♦ 

EXAMPLE 5 Your intuition should lead you to suspect that a flow line of the 
vector field F(x, y) = — yi + xj should be circular as showninFigure 3.36. Indeed 
if x: [0, 2jt) — > R 2 is given by x(t) = (a cos t , a sin t), where a is constant, then 

x'(f) = — a sinf i + a cost j = F(a cos/, a sin?), 

so such paths are indeed flow lines. 

Finding all possible flow lines of F(x , y) = — yi + xj is a more involved task. 
If x(/) = (x(f), y(t)) is a flow line, then, by Definition 3.2, we must have 

At) = x'(t)i + y'(t)j = - y (0i + *(0j = F(x(0). 
Equating components, 

\x'{t) = -y(t) 
y'(t) = x(t) ■ 

This is an example of a first-order system of differential equations. It turns out 
that all solutions to this system are of the form 



x(t) = (a cos t — b sin?, a sin? + b cos ?), 



Chapter 3 | Vector-Valued Functions 



where a and b are arbitrary constants. It's not difficult to see that such paths trace 
circles when at least one of a or b is nonzero. ♦ 

In general, if F is a vector field on R", finding the flow lines of F is equivalent 
to solving the first-order system of differential equations 

x[(t) = FiCti(f), X 2 (t), X„(t)) 
x' 2 (t) = F 2 (xi(t), x 2 (t), x„(t)) 

. X'„(t) = F n (Xi(t), X 2 (t), . . . , Xn(t)) 

for the functions X\(t), . . . , x n (t) that are the components of the flow line x. (The 
function F, is just the ith component function of the vector field F.) Such a 
problem takes us squarely into the realm of the theory of differential equations, a 
fascinating subject, but not of primary concern at the moment. 



3.3 Exercises 



In Exercises 1-6, sketch the given vector fields on R 2 . 



1. 


F 


= yi- X \ 


2. 


F 


= xi- yj 


3. 


F 


= {-*,y) 


4. 


F 


= (x,x 2 ) 


5. 


F 


= (x 2 ,x) 


6. 


F 


= (y\y) 



In Exercises 7-12, sketch the given vector field on R 3 . 

7. F = 3i + 2j + k 

8. F = (y, -x,0) 

9. F = (0, z, -y) 

10. F = (y, -x,2) 

11. F = (y,-x,z) 

V x 

12. F= - y = i — = j 

V-* 2 + y 2 + z 2 V-* 2 + y 2 + z 2 

7 

+ ; = =k 

s/x 2 + y 2 + z 2 

In Exercises 13-16, use a computer to plot the given vector 
fields over the indicated ranges. 

^ 13. F = (x - y, x + y); -1 < x < 1, -1 < y < 1 
^14. F = (y 3 x, x 2 y); -2 < x < 2, -2 < y < 2 

15. F = (x siny, y cosx); —2jt<x<2jt, 
—2it < y < 2tc 

^ 16. F = (cos(x - y), sin(x + y)); -2n < x < 2jt, 
—2it < y < 2tc 



In Exercises 17-19, verify that the path given is a flow line of 
the indicated vector field. Justify the result geometrically with 
an appropriate sketch. 

17. x(t) = (sinf, cosr, 0), F = (y, -x, 0) 

18. x(r) = (sinf, cos?, 2t), F = (y, -x, 2) 

19. x(r) = (sinf, cosf, e 2t ), F = (y, -x, 2z) 

In Exercises 20-22, calculate the flow line x(f ) of the given 
vector field F that passes through the indicated point at the 
specified value of t. 

20. F(x,y) = -xi + yj; x(0) = (2, 1) 

21. F(x,y) = (x 2 ,y); x(l) = (l,e) 

22. F(x,y,z) = 2i-3yj + Z 3 k; x(0) = (3,5,7) 

23. Consider the vector field F = 3 i - 2 j + k. 

(a) Show that F is a gradient field. 

(b) Describe the equipotential surfaces of F in words 
and with sketches. 

24. Consider the vector field F = 2x i + 2y j — 3k. 

(a) Show that F is a gradient field. 

(b) Describe the equipotential surfaces of F in words 
and with sketches. 

25. If x is a flow line of a gradient vector field F = V/, 
show that the function G(f ) = /(x(f )) is an increasing 
function of f. (Hint: Show that G'(t) is always non- 
negative.) Thus, we see that a particle traveling along 
a flow line of the gradient field F = V/ will move 
from lower to higher values of the potential function 
/. That's why physicists define a potential function of 
a gradient vector field F to be a function g such that 
F = — Vg (i.e., so that particles traveling along flow 
lines move from higher to lower values of g). 



3.4 | Gradient, Divergence, Curl, and the Del Operator 227 



LetF: X C R" — > R" be a continuous vector field. Let (a, b)be 
an interval in "Rthat contains 0. (Think of {a, b)asa "timeinter- 
val") A flow ofF is a differentiable function cp: X x (a, b) — > 
R" ofn + 1 variables such that 



dt 



0(x, t) = F(0(x, f)); 0(x, 0) = x. 



Intuitively, we think of 0(x, t) as the point at time t on the flow 
line ofF that passes through x at time 0. (See Figure 3.37.) 
Thus, the flow ofF is, in a sense, the collection of all flow lines 
ofF. Exercises 26-31 concern flows of vector fields. 



F(0(x, 0) 
<Hx,f)/ 




Figure 3.37 The flow of the vector field F. 



26. Verify that 

0:R 2 x R^ R 2 , 

<j>(x, .v, t) 



x + y t , x ~y - t 

e -\ e , 

2 2 



x + y , y — x 

e' + : e~ 

2 2 

is a flow of the vector field F(x, y) = (y, x). 



27. Verify that 
</>:R 2 x R^ R 2 , 

4>(x, y, t) = (y sinf + x cos t, y cos? — x sinf) 
is a flow of the vector field F(x, y) = (y, —x). 

28. Verify that 
0:R 3 x R^ R 3 , 

4>(x, y, z, t) = (x cos2f — y sin 2;, y cos2f 
+ x sin2f, ze~') 



is a flow of the vector field F(x,y,z) 
2x j — z k. 



-2vi + 



29. Show that if <j>: X x (a, fo) R" is a flow ofF, then, 
for a fixed point xo in X, the map x: (a, b) — ► R" given 
by x(f) = 0(xo, f) is a flow line ofF. 

30. If 0 is a flow of the vector field F, explain why 
0(0(x, r), s) = 0(x, s + r). (Hint: Relate the value of 
the flow 4> at (x, t) to the flow line ofF through x. You 
may assume the fact that the flow line of a continuous 
vector field at a given point and time is determined 
uniquely.) 

31 . Derive the equation of first variation for a flow of a 
vector field. That is, if F is a vector field of class C 1 
with flow <p of class C 2 , show that 



dt 



D x 0(x, 0 = DF(0(x, t))D %< p(x, t). 



Here the expression "D x 0(x, t)" means to differentiate 
(p with respect to the variables x\ , X2, . . . , x„, that is, 
by holding t fixed. 



3.4 Gradient, Divergence, Curl, and the Del 
Operator 

In this section, we consider certain types of differentiation operations on vector 
and scalar fields. These operations are as follows: 

1. The gradient, which turns a scalar field into a vector field. 

2. The divergence, which turns a vector field into a scalar field. 

3. The curl, which turns a vector field into another vector field. (Note: The curl 
will be denned only for vector fields on R 3 .) 

We begin by defining these operations from a purely computational point of view. 
Gradually, we shall come to understand their geometric significance. 



The Del Operator 

The del operator, denoted V, is an odd creature. It leads a double life as both 
differential operator and vector. In Cartesian coordinates on R 3 , del is defined by 



Chapter 3 | Vector-Valued Functions 

the curious expression 




The "empty" partial derivatives are the components of a vector that awaits suitable 
scalar and vector fields on which to act. Del operates on (i.e., transforms) fields 
via "multiplication" of vectors, interpreted by using partial differentiation. 

For example, if /: X c R 3 — >• R is a differentiable function (scalar field), 
the gradient of / may be considered to be the result of multiplying the vector V 
by the scalar /, except that when we "multiply" each component of V by /, we 
actually compute the appropriate partial derivative: 

/ 3 3 3 \ df df df 

Vf(x,y,z)= i— +j— +k- )f(x,y,z)= -M+/j + /k. 

\ dx ay dz ) dx dy dz 

The del operator can also be defined in R", for arbitrary n. If we take 
x\, X2, ■ ■ ■ , x n to be coordinates for R", then del is simply 




where e, ■ = (0, . . . , 1, . . . , 0), i = 1, . . . , n, is the standard basis vector for R". 
The Divergence of a Vector Field 



Whereas taking the gradient of a scalar field yields a vector field the process of 
taking the divergence does just the opposite: It turns a vector field into a scalar 
field. 



DEFINITION 4.1 Let F: X C R" R" be a differentiable vector field. 
Then the divergence of F, denoted div F or V • F (the latter read "del dot 
F"), is the scalar field 

dFi dF 2 dF n 
div F = V • F = — L + — L + ... + — 1, 

3xi dxi dx„ 

where Cartesian coordinates for R" and F\ , . . . , F„ are the 

component functions of F. 



It is essential that Cartesian coordinates be used in the formula of Definition 4.1. 
(Later in this section we shall see what div F looks like in cylindrical and spherical 
coordinates for R 3 .) 

EXAMPLE 1 IfF = jr 2 vi + x Z j+xyzk,then 

3 3d 
div F = — (x 2 y) + — (jcz) + — (xyz) = 2xy + 0 + xy = 3xy. ♦ 
dx dy dz 



3.4 | Gradient, Divergence, Curl, and the Del Operator 229 



The notation for the divergence involving the dot product and the del operator 
is especially apt: If we write 



F = Fid + F 2 e 2 H h F n e n , 



then, 



/ 3 3 3 \ , 

V-F= ei— +e 2 — + --- + e„— ■ (Fid + F 2 e 2 + ■ ■ ■ + F„e„) 

\ dXl 0X2 ox n J 

_ dFi dF 2 dF n 
dx\ 9x2 3x„ ' 

where, once again, we interpret "multiplying" a function by a partial differential 
operator as performing that partial differentiation on the given function. 

Intuitively, the value of the divergence of a vector field at a particular point 
gives a measure of the "net mass flow" or "flux density" of the vector field in 
or out of that point. To understand what such a statement means, imagine that 
the vector field F represents velocity of a fluid. If V • F is zero at a point, then 
the rate at which fluid is flowing into that point is equal to the rate at which 
fluid is flowing out. Positive divergence at a point signifies more fluid flowing out 
than in, while negative divergence signifies just the opposite. We will make these 
assertions more precise, even prove them, when we have some integral vector 
calculus at our disposal. For now, however, we remark that a vector field F such 
that V • F = 0 everywhere is called incompressible or solenoidal. 

EXAMPLE 2 The vector field F = xi + yj has 

V-F= A (je )+ ^(y) = 2. 
dx ay 

This vector field is shown in Figure 3.38. At any point in R 2 , the arrow whose 
tail is at that point is longer than the arrow whose head is there. Hence, there is 
greater flow away from each point than into it; that is, F is "diverging" at every 
point. (Thus, we see the origin of the term "divergence.") 

The vector field G = — xi — yj points in the direction opposite to the vector 
field F of Figure 3.38 (see Figure 3.39), and it should be clear how G's divergence 
of —2 is reflected in the diagram. ♦ 



EXAMPLE 3 The constant vector field F(x, y, z) = a shown in Figure 3.40 
is incompressible. Intuitively, we can see that each point of R 3 has an arrow 
representing a with its tail at that point and another arrow, also representing a, 
with its head there. 

The vector field G = yi — xj has 

v-g= A (v) + A ( _ x) = o. 

ax ay 

A sketch of G reveals that it looks like the velocity field of a rotating fluid, without 
either a source or a sink. (See Figure 3.41 .) ♦ 



The Curl of a Vector Field 

If the gradient is the result of performing "scalar multiplication" with the del 
operator and a scalar field, and the divergence is the result of performing the 
"dot product" of del with a vector field, then there seems to be only one simple 



Chapter 3 | Vector-Valued Functions 




Figure 3.40 The constant vector 
field F = a. 



Figure 3.41 The vector field 
G = yi — x'] resembles the 
velocity field of a rotating fluid. 



differential operation left to be built from del. We call it the curl of a vector field 
and define it as follows: 



DEFINITION 4.2 Let F: X c R 3 -> R 3 be a differentiable vector field on 
R 3 only. The curl of F, denoted curl F or V x F (the latter read "del cross 
F"), is the vector field 

(3 3 3 \ 

V + V + V ) x + F ^ + F3k > 
dx ay dz ) 



i j k 

d/dx d/dy d/dz 
F\ F 2 F 3 



^EL - i + ( ?H _ ^fl\ ■ + ( ^fl _ | 

dy dz J \ dz dx J \ dx dy 



There is no good reason to remember the formula for the components of the 
curl — instead simply compute the cross product explicitly. 

EXAMPLE 4 If F = x 2 yi - 2xzj + (x + y - z)k, then 



V x F 



i j k 

d/dx d/dy d/dz 

— 2xz x + y — z 



x 2 y 



3 3 \ ( 3 3 

— (x + y-z)- ^~ 2xz) ) 1 + \dz~ (x2y) ~ 9^ (X + y ~ Z - 



3 



9 



= (1 + 2x)i - j - (x 2 + 2z)k. 



3.4 | Gradient, Divergence, Curl, and the Del Operator 231 




Figure 3.42 A twig in a pond where water moves with velocity given by a vector field F. In the left figure, the twig does 
not rotate as it travels, so curl F = 0. In the right figure, curl F / 0, since the twig rotates. 



One would think that, with a name like "curl," V x F should measure how 
much a vector field curls. Indeed the curl does measure, in a sense, the twisting 
or circulation of a vector field but in a subtle way: Imagine that F represents the 
velocity of a stream or lake. Drop a small twig in the lake and watch it travel. 
The twig may perhaps be pushed by the current so that it travels in a large circle, 
but the curl will not detect this. What curl F measures is how quickly and in what 
orientation the twig itself rotates as it moves. (See Figure 3.42.) We prove this 
assertion much later, when we know something about line and surface integrals. 
For now, we simply point out some terminology: A vector field F is said to be 
irrotational if V x F = 0 everywhere. 



EXAMPLE 5 Let F = (3x 2 z + y 2 ) i + 2xy j + (x 3 - 2z) k. Then 



V x F 



l 

d/dx 
3x 2 z + y 2 

a 



j 

d/dy 
2xy 



k 

d/dz 



2z 



dy 



(x> - 2z) 



^(2xy)W(^(3x 2 z + y 2 ) 



dx 



(x 3 -2z) 



dx 



(2xy) 



3 

9y 



(3* 2 z + y 2 ) I k 



= (0 - 0)i + (3x 2 - 3x 2 )\ + (2y - 2y)k = 0. 



Thus, F is irrotational. 



Two Vector-analytic Results 

It turns out that the vector field F in Example 5 is also a gradient field. Indeed 
F = V/, where f(x, y, z) = x 3 z + xy 2 — z 2 . (We'll leave it to you to verify 
this.) In fact, this is not mere coincidence but an illustration of a basic result 
about scalar-valued functions and the del operator: 



THEOREM 4.3 Let /: X C R 3 -> R be of class C 2 . Then curl (grad/) = 0. 
That is, gradient fields are irrotational. 



Chapter 3 | Vector-Valued Functions 



PROOF Using the del operator, we rewrite the conclusion as 



V x (V/) = 0, 

which might lead you to think that the proof involves nothing more than noting 
that V/ is a "scalar" times V, hence, "parallel" to V, so thatthe cross productmust 
be the zero vector. However, V is not an ordinary vector, and the multiplications 
involved are not the usual ones. A real proof is needed. 

Such a proof is not hard to produce: We need only start calculating V x (V/). 
We have 



df df df 
dx dy dz 



Therefore, 



V x (V/) = 



i j k 

d/dx d/dy d/dz 
df/dx df/dy df/dz 



d 2 f 

dydz dzdy 



dzdy) \ 



d 2 f d 2 f 



dzdx dxdz 



j + 



d 2 f d 2 f 



dxdy dydx 



Since / is of class C 2 , we know that the mixed second partials don't depend 
on the order of differentiation. Hence, each component of V x (V/) is zero, as 
desired. ■ 

There is another result concerning vector fields and the del operator that is 
similar to Theorem 4.3: 



THEOREM 4.4 Let F:XcR 3 ^R 3 be a vector field of class C 2 . Then 
div (curl F) = 0. That is, curl F is an incompressible vector field. 



The proof is left to you. 

EXAMPLE 6 If F = (xz — e 2x cos z) i - yz j + e 2l (sin y + 2 sin z) k, then 

3 3 3 

V • F = — (xz — e 2x cosz) + — (-yz) + — (e 2x (smy + 2sinz)) 
dx dy dz 

= z - 2e 2x cos z - z + 2e 2x cos z = 0 

for all (ij,z)£R 5 . Hence, F is incompressible. We'll leave it to you to check 
that F = V x G, where G(x, y, z) = e 2x cos y i + e 2x sin z j + xyz k, so that, in 
view of Theorem 4.4 the incompressibility of F is not really a surprise. ♦ 



Other Coordinate Formulations (optional) 

We have introduced the gradient, divergence, and curl by formulas in Cartesian 
coordinates and have, at least briefly, discussed their geometric significance. Since 
certain situations may necessitate the use of cylindrical or spherical coordinates, 
we next list the formulas for the gradient, divergence, and curl in these coordinate 
systems. Before we do, however, a remark about notation is in order. Recall that 
in cylindrical coordinates, there are three unit vectors e r , e$, and e z that point in 
the directions of increasing r, 6, and z coordinates, respectively. Thus, a vector 



3.4 | Gradient, Divergence, Curl, and the Del Operator 233 



field F on R 3 may be written as 

F = F r e r + F g e e + F-e.. 

In general, the component functions F r , Fg, and F z are each functions of the 
three coordinates r, 9, and z; the subscripts serve only to indicate to which of 
the vectors e, , eg, and e ; that particular component function should be attached. 
Similar comments apply to spherical coordinates, of course: There are three unit 
vectors e p , e 9 , and eg, and any vector field F can be written as 

F = F p e p + F v e v + F e e g . 



THEOREM 4.5 Let /: X c R 3 -> R and F: Y C R 3 

scalar and vector fields, respectively. Then 



3/ 13/ , 3/ 

— e r H e e H e,; 

9r r 30 3z 



divF = 



curl F = 



3 3F 0 3 

or 39 dz 



R be differentiate 

(3) 
(4) 



e r rt 0 

3/3r 3/36* 
F r rFg 



e 

3/3z 



(5) 



PROOF We'll prove formula (4) only, since the argument should be sufficiently 
clear so that it can be modified to give proofs of formulas (3) and (5). The idea is 
simply to rewrite all rectangular symbols in terms of cylindrical ones. 
From the equations in (8) of § 1 .7, we have 



e,- = cos 6 i + sin 9 j 
eg = — sin 9 i + cos 6 j . 
e- = k 



(6) 



From the chain rule, we have the following relations between rectangular and 
cylindrical differential operators: 

3 3 3 

— = cos 9 h sin 6 — 

3r dx 3y 

3 3 3 

— = —rsm9 h r cos 9 — . 

89 dx 3y 

3 _ 3 

dz dz 

These relations can be solved algebraically for d/dx, d/dy, and 3/3z to yield 



a 


= cos 


3 

9 


sin 9 


3 


37 


dr 


r 


39 


3 


= sin 6 


3 

Vr + 


cos 9 


8 


3v" 


r 


39 


3 


3 








dz 


~ d~Z 









(7) 



234 Chapter 3 I Vector- Valued Functions 



Hence, we can use (6) and (7) to rewrite the expression for the divergence of a 
vector field on R 3 : 



V-F 



3 3 3 

IT 1 + IT J + IT k ) * ( F '" e '- + FsCs + F ^ 
dx dy dz 



1 cos 



,1 

dr 



sin# 3 
~r r ~d~9 



j sin (9 



3 cos9 3 



8r 



r 3(9 



dz 



[(F r cos 9 - F e sin 9) i + (F r sin <9 + F e cos 0) j + F z k]. 



(We used the equations in (7) to rewrite the partial operators d/dx, 3/3 y, and 
3/3z appearing in del and the equations in (6) to replace the cylindrical basis 
vectors e r , eg, and e z by expressions involving i, j, and k.) Performing the dot 
product and using the product rule yields 



V-F 



cost 



3 sva9 3 

i 

dr r d9 

3 cos# 3 

+ |sin6> 1 

dr r d9 



(F r cos 9 — F g sin#) 



(F r sin 9 + F e cos 0) H F- 

dz 



dF r 3 

= cost/ [cos 9 h F r — (cos#) 

dr dr 



cos v sin 



3r 

dF r 



dr 



sin9 
r 

sin 



+ sm9 \ sm9 — - + F r — (sm.9) 
dr dr 

dF 0 3 

+ sin# | cosf h F g — (cos$) 

dr dr 



dF r 3 

cos 9 h F r — (cos 9) 

3(9 3(9 



r 

cos 9 
r 

cos 9 



sin 9- 



dF e 
d9 



3 

F e —(sin9) 
3(9 v 



dF r d 
sm9 — - + F r — (sine) 
36> 89 



cos9- 



d_F^ 
89 



3 

Fa — (cos 9) 
d9 



+ 



dz 



After some additional algebra, we find that 



3 F 

V • F = (cos 2 9 + sin 2 9) — - + 

dr 



sin 2 9 + cos 2 ( 



\F r 



+ 

dF r 



sin 2 6> + cos 2 9 \ dF e 



36 



dF z 
IF 



1 1 dF e 8F, 
dr r r d9 dz 



( 



\ 



r h F r H rF, 

3r 3(9 3z V 



as desired. 



In spherical coordinates, the story for the gradient, divergence, and curl is 
more complicated algebraically, although the ideas behind the proof are essentially 



3.4 I Exercises 235 



the same. We state the relevant results and leave to you the rather tedious task of 
verifying them. 



THEOREM 4.6 Let /: X C R 3 -> R and F: Y C R 3 -> R 3 be differentiable 
scalar and vector fields, respectively. Then the following formulas hold: 



v/ = 




1 df 
e 

p d<p 


1 

p sin<p 


de ee; 


V-F = 


1 3 , 

p L dp 


%) + 


1 3 
p sin<p 3^> 


(sin<p Fy) + 




1 


e P 




p sin <p e# 


V x F = 


d/dp 


d/dcp 


3/36* 


p 2 sin <p 










p sin F g 



1 



3F e 



p sin<p dO 



(8) 
(9) 

(10) 



3.4 Exercises 



Calculate the divergence of the vector fields given in Exer- 
cises 1-6. 



1. F 

3. F 

4. F 

5. F 

6. F 



x 2 i + y 2 ] 



2. F 



y 2 i + x 2 } 



(x + y)i + (y + z)j + (jc + z)k 
Z cos (e r ) i + xy/z 2 + 1 j + e 2v sin 3jc k 

Xjei + 2x|e 2 H h «x 2 e„ 

xiei + 2xie2 + • • • + nx\e„ 
Find the curl of the vector fields given in Exercises 7-11. 

7. F = x 2 i — xe y j + 2xyz k 

8. F = xi + yj + zk 

9. F = (x + yz)i + (y + xz)} + (z + xy)k 

1 0. F = (cos yz — x)i + (cos xz — y)j + (cos xy — z)k 

11. F = y 2 zi + e xyz j+x 2 yk 

12. (a) Consider again the vector field in Exercise 8 and 

its curl. Sketch the vector field and use your pic- 
ture to explain geometrically why the curl is as you 
calculated. 

(b) Use geometry to determine V x F, where F = 
(xi + yj + zk) 

y/x 2 + y 2 + z 2 ' 

(c) For F as in part (b), verify your intuition by explic- 
itly computing V x F. 

13. Can you tell in what portions of R 2 , the vector fields 
shown in Figures 3.43-3.46 have positive divergence? 
Negative divergence? 



\ \ \ \ i i t 

\ \ \ \ i i i 

w \ u I I 

v x \ \ \ \ I 



/ / I 1 I i \ 

/ / I I I I I 

////Ml 



1 1 t t t / / 
III//// 
I / / t f / / 

I / / / / S S 



I \ \ \ \ \ N 
I I \ \ \ \ \ 
\ M \ \ \ \ 



Figure 3.43 Vector field for Exercise 13(a). 



s \ \ \ \ 



i i / / / 



S S / / / / \ \ \ \ *s ^ ^ 



Figure 3.44 Vector field for Exercise 13(b). 



Chapter 3 | Vector-Valued Functions 



Figure 3.45 Vector field for Exercise 13(c). 



////II I 
////ill 
///iii i 
✓ ✓///// 



X \ \ \ \ \ \ 
S \ \ \ \ \ | 
X \ \ \ \ \ ( 

\\ \ \ \ u 



M \ \ V \ s 

\ \ \ \ \ \ \ 

\ V \ \ \ V V 

\ S \ \ \ \ N 



III//// 
I I I I / / / 
III//// 
III//// 



Figure 3.46 Vector field for Exercise 13(d). 

14. Check that if f(x, y,z) = x 2 sin y + y 2 cosz, then 

V x (V/) = 0. 

15. Check that if F(x, y, z) = xyzi — e z cos xj + xy 2 z 3 K 
then 

V • (V x F) = 0. 

16. Prove Theorem 4.4. 

In Exercises 1 7-20, let r = x i + y j + zkand let r denote ||r||. 
Verify the following: 

17. Vr" =nr"- 2 r 

18. V(lnr)= ^ 

r i 

19. V • (r"r) = (n + 3)r" 

20. V x (r n r) = 0 

In Exercises 21—25, establish the given identities. (You may 
assume that any functions and vector fields are appropriately 
differentiable.) 

21. V-(F + G) = V-F + V-G 



22. V x (F + G) = V x F + V x G 

23. V.(/F) = /V.F + F.V/ 

24. V x (/F) = /V x F + V/ x F 

25. V-(FxG) = G- VxF-F-VxG 

26. Prove formulas (3) and (5) of Theorem 4.5. 

27. Establish the formula for the gradient of a function in 
spherical coordinates given in Theorem 4.6. 

28. The Laplacian operator, denoted V 2 , is the second- 
order partial differential operator defined by 

, s 2 s 2 a 2 

V 2 = 1 1 . 

dx 2 3y 2 dz 2 

(a) Explain why it makes sense to think of V 2 as V • V . 

(b) Show that if / and g are functions of class C 2 , then 

V 2 (/g) = fV 2 g + gV 2 f + 2(V/ • Vg). 

(c) Show that 

V-(/Vg-gV/) = /V 2 g-gV 2 /. 

29. Show that V • (/V/) = || V/|| 2 + /V 2 /. 

30. Show that V x (V x F) = V(V • F) - V 2 F. (Here 
V 2 F means to take the Laplacian of each component 
function of F.) 

Let X be an open set in R", F:XC R" R" a vector field 
on X, and a € X. If y is any unit vector in R", we define the 
directional derivative of 'F at a in the direction of v, denoted 
D v F(a), by 

1 

D v F(a) = lim -(F(a + hv) - F(a)), 

h^O h 

provided that the limit exists. Exercises 31-34 involve direc- 
tional derivatives of vector fields. 

31. (a) In analogy with the directional derivative of a 

scalar-valued function defined in §2.6, show that 



D v F(a) 



dt 



F(a + 1 v) 



[=0 



(b) Use the result of part (a) and the chain rule to show 
that, if F is differentiable at a, then 

D v F(a) = DF(a)v, 

where v is interpreted to be an n x 1 matrix. (Note 
that this result makes it straightforward to calculate 
directional derivatives of vector fields.) 

32. Show that the directional derivative of a vector field 
F is the vector whose components are the directional 



Miscellaneous Exercises for Chapter 3 237 



derivatives of the component functions F\, . . . , F„ of 
F, that is, that 

D v F(a) = (DyFii*), D v F 2 (a), D v F n (a)). 

33. LetF = vzi + xzj +xyk.FindD (i _ j+k)/V 3F(3,2, 1). 
(Hint: See Exercise 31.) 



34. Let F = x i + y j + z k. Show that D v F(a) = v for any 
point a € R 3 and any unit vector veR 3 . More gener- 
ally, if F = (x\, X2, ■ ■ ■ , x n ), a = (fli, ai, . . . , a n ), and 
v = (t>i, V2 v n ), show that _D v F(a) = v. 



True/False Exercises for Chapter 3 



1 . If a path x remains a constant distance from the origin, 
then the velocity of x is perpendicular to x. 

2. If a path is parametrized by arclength, then its velocity 
vector is constant. 

3. If a path is parametrized by arclength, then its velocity 
and acceleration are orthogonal. 

' ■x(f)|[ = ||x'(f)[|. 



4. 



dt 
d 



5. — (x x y) 

dt K 3 



6. K 



dT 

dt 



dy dx 

xx h y x — . 

dt J dt 



7. |T| 



dB 

ds 



8. The curvature k is always nonnegative. 

9. The torsion r is always nonnegative. 

10. N 



dT 

ds 



11. If a path x has zero curvature, then its acceleration is 
always parallel to its velocity. 

12. If a path x has a constant binormal vector B, then r =0. 



|a(r)[| 2 



14. grad / is a scalar field. 

15. div F is a vector field. 

16. curl F is a vector field. 



1 7. grad(div F) is a vector field. 

18. div(curl(grad /)) is a vector field. 

1 9. grad/ x div F is a vector field. 

20. The path x(f) = (2 cos t, 4 sinr, t) is a flow line of the 

y 

vector field F(x ,y, z) = — — i + 2x j + z k. 

21. The path x(?) = (e' cosf , e'(cos t + sinf), e' sin?) is a 
flow line of the vector field F(jc, y, z) = (x — z)i + 
2jc j + y k. 

22. The vector field F = 2xy cos z i — y 2 cos z j + e xy k is 
incompressible. 

23. The vector field F = 2xy cos z i — y 2 cos z j + e xy k is 
irrotational. 

24. V x (V/) = 0 for all functions /: R 3 — » R. 

25. If V • F = 0 and V x F = 0, then F = 0. 

26. V • (F x G) = F • (V x G) + G • (V x F). 

27. If F = curl G, then F is solenoidal. 

28. The vector field F = 2x sin y cos z i + x 2 cos y cos z 
j + x 2 sin y sin z k is the gradient of a function / of 
class C 2 . 

29. There is a vector field F of class C 2 on R 3 such that 
V x F = x cos 2 y i + 3y j — xyz 2 k. 

30. If F and G are gradient fields, then F x G is incom- 
pressible. 



Miscellaneous Exercises for Chapter 3 



1 . Figure 3.47 shows the plots of six paths x in the plane. 
Match each parametric description with the correct 
graph. 

(a) x(/) = (s'mlt, sin 3i) 

(b) x(f) = (t + sin5f, t 2 + cos 6?) 

(c) x(r) = (f 2 + l,f 3 -?) 

(d) x(f) = (2? + sin4f, t - sin5f) 

(e) x(t) = {t -t 2 ,t 3 -t) 

(f) x(t) = (sin(? + sin 30, cosf) 



2. Figure 3.48 shows the plots of six paths x in R 3 . Match 
each parametric description with the correct graph. 

(a) x(f) = (t + cos3f, t 2 + sin5f, sin At) 

(b) x(f) = (2 cos 3 1, 3 sin 3 1, cos2f) 

(c) x(0 = (15 cos /, 23 sin t , At) 

(d) x(0 = (cos3r, cos5f, sin4f) 

(e) x(f) = (2f cosf, 2f sin?, 4f) 

(f) x(f) = (r 2 + l,? 3 -f,? 4 -r 2 ) 



238 Chapter 3 | Vector-Valued Functions 




Miscellaneous Exercises for Chapter 3 239 



3. Suppose that x is a C 2 path with nonzero velocity. Show 
that x has constant speed if and only if its velocity and 
acceleration vectors are always perpendicular to one 
another. 

4. You are at Vertigo Amusement Park riding the new 
Vector roller coaster. The path of your car is given by 



x(?) = (x(t), y{t)), where 



x(f) 



t/60 1Tt t/60 ■ Jtt 

e ' cos — ,e ' sm — , 
30 30 

2r(10 - t)(t - 90) 2 



80 + 



10 6 



where / = 0 corresponds to the beginning of your 
three-minute ride, measured in seconds, and spatial 
dimensions are measured in feet. It is a calm day, but 
after 90 sec of your ride your glasses suddenly fly off 
your face. 

(a) Neglecting the effect of gravity, where will your 
glasses be 2 sec later? 

(b) What if gravity is taken into account? 

5. Show that the curve traced parametrically by 

1 



x(f) = ^cos(f - 1), t - 1, 

istangenttothesurface.ii; 3 + y 3 + z 3 
t = 1. 



xyz = 0 when 



6. Gregor, the cockroach, is on the edge of a Ferris wheel 
that is rotating at a rate of 2 rev/min (counterclock- 
wise as you observe him). Gregor is crawling along 
a spoke toward the center of the wheel at a rate of 
3 in/min. 

(a) Using polar coordinates with the center of the 
wheel as origin, assume that Gregor starts (at time 
t = 0) at the point r = 20 ft, 6 = 0. Give paramet- 
ric equations for Gregor's polar coordinates r and 
6 at time t (in minutes). 

(b) Give parametric equations for Gregor's Cartesian 
coordinates at time t. 

(c) Determine the distance Gregor has traveled once 
he reaches the center of the wheel. Express your 
answer as an integral and evaluate it numerically. 

If you have used a drawing program on a computer, you have 
probably worked with a curve known as a Bezier curve. 1 Such 
a curve is defined parametrically by using several control 
points in the plane to shape the curve. In Exercises 7-12, 
we discuss various aspects of quadratic Bezier curves. These 
curves are defined by using three fixed control points {x\ , y\), 
( x 2, yi), and(xi, yi) and a nonnegative constant w. The Bezier 
curve defined by this information is given by x: [0, 1] — > R 2 , 



X(t): 

y(t) : 



(1 - tfx\ + 2wt(l - t)x 2 + t 2 X } 
(\-tf + 2wt(\-t) + t 2 

(1 -tf yi +2wtQ_ -r)y 2 + f 2 y 3 
(1 -t) 2 + 2wt{\ -t) + t 2 



0 < t < 1. 



(1) 

^ 7. Let the control points be (1,0), (0, 1), and (1, 1). 
Use a computer to graph the Bezier curve for w = 
0, 1/2, 1, 2, 5. What happens as w increases? 

8. Repeat Exercise 7 for the control points (—1, — 1), 
(1,3), and (4, 1). 

9. (a) Show that the Bezier curve given by the paramet- 

ric equations in ( 1 ) has (x\ , y\ ) as initial point and 
(x3, V3) as terminal point. 

(b) Show that x(i) lies on the line segment joining 
(*2, yi) to the midpoint of the line segment joining 
(x\,y\) to (x 3 , y 3 ). 

10. In general the control points (xi,y{), (x2,y2), and 
(*3 , )?3 ) will form a triangle, known as the control poly- 
gon for the curve. Assume in this problem that w > 0. 
By calculating x'(0) and x'(l), show that the tangent 
lines to the curve at x(0) and x(l) intersect at (X2, V2). 
Hence, the control triangle has two of its sides tangent 
to the curve. 

11. In this problem, you will establish the geometric sig- 
nificance of the constant w appearing in the equations 
in(l). 

(a) Calculate the distance a between x( j ) and (x2 , ^2). 

(b) Calculate the distance b between x(^) and the 
midpoint of the line segment joining (xi, y\) and 

C*3, yj)- 

(c) Show that w = b/a. By part (b) of Exercise 9, 
x(|) divides the line segment joining (X2, 3^) to 
the midpoint of the line segment joining (x\, y\) 
to (*3, V3) into two pieces, and w represents the 
ratio of the lengths of the two pieces. 

12. Determine the Bezier parametrization for the portion 
of the parabola y = x 2 between the points (—2, 4) and 
(2, 4) as follows: 

(a) Two of the three control points must be (—2, 4) and 
(2, 4). Find the third control point using the result 
of Exercise 10. 

(b) Using part (a) and Exercise 9, we must have that 
x(i) lies on the y-axis and, hence, at the point 



2 P. Bezier was an automobile design engineer for Renault. See D. Cox, J. Little, and D. O'Shea, Ide- 
als, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative 
Algebra, 3rd ed. (Springer- Verlag, New York, 2007), pp. 28-29. Exercises 7-1 1 adapted with permission. 



240 Chapter 3 | Vector- Valued Functions 



(0, 0). Use the result of Exercise 11 to determine 
the constant w. 

(c) Now write the Bezier parametrization. You should 
be able to check that your answer is correct. 

1 3. Let x: (0, 7t) -*■ R 2 be the path given by 

x(?) = (sin?, cos? + In tan |) , 

where ? is the angle that the y-axis makes with the 
vector x(?). The image of x is called the tractrix. (See 
Figure 3.49.) 

(a) Show that x has nonzero speed except when ? = 
n/2. 

(b) Show that the length of the segment of the tangent 
to the tractrix between the point of tangency and 
the y-axis is always equal to 1. This means that 
the image curve has the following description: Let 
a horse pull a heavy load by a rope of length 1 . 



of class C 2 . Use equation (17) in §3.2 to derive the 
curvature formula 




Figure 3.49 The tractrix 
of Exercise 13. 

Suppose that the horse initially is at (0, 0), the load 
at (1, 0), and let the horse walk along the v-axis. 
The load follows the image of the tractrix. 

14. Another way to parametrize the tractrix path given in 
Exercise 13 is 

y: (-oo, O)-*- R 2 , 



where y(r) 



O-P 



dp 



(a) Show that y satisfies the property described in part 
(b) of Exercise 13. 

(b) In fact, y is actually a reparametrization of part of 
the path x of Exercise 13. Without proving this fact 
in detail, indicate what portion of the image of x 
the image of y covers. 

15. Suppose that a plane curve is given in polar coordi- 
nates by the equation r = f{6), where / is a function 



K(9) 



' + 2r' : 



( r 2 _|_ r /2)3/2 



(Hint: First give parametric equations for the curve in 
Cartesian coordinates using 9 as the parameter.) 

16. Use the result of Exercise 15 to calculate the curvature 
of the lemniscate r 2 = cos 26. 

Let x: / — > R 2 be a path of class C 2 that is not a straight line 
and such that x'(?) ^ 0. Choose some to € I and let 

y(?) = x(?) - s(f)T(f), 

where s(t) = f' ||x'(r)|| dr is the arclength function and T is 
the unit tangent vector. The path y: / —> R 2 is called the invo- 
lute of x. Exercises 1 7—19 concern involutes of paths. 

17. (a) Calculate the involute of the circular path of radius 

a, that is, x(?) = (a cos ?, a sin ?). (Take ?o to be 0.) 

(b) Let a = 1 and use a computer to graph the path x 
and the involute path y on the same set of axes. 

1 8. Show that the unit tangent vector to the involute at ? 
is the opposite of the unit normal vector N(?) to the 
original path x. (Hint: Use the Frenet-Serret formulas 
and the fact that a plane curve has torsion equal to zero 
everywhere.) 

19. Show that the involute y of the path x is formed by 
unwinding a taut string that has been wrapped around 
x as follows: 

(a) Show that the distance in R 2 between a point x(?) 
on the original path and the corresponding point 
y(?) on the involute is equal to the distance traveled 
from x(?o) to x(?) along the underlying curve of x. 

(b) Show that the distance between a point x(?) on the 
path and the corresponding point y(?) on the in- 
volute is equal to the distance from x(?) to y(?) 
measured along the tangent emanating from x(?). 
Then finish the argument. 

Let x: / — > R 2 be a path of class C 2 that is not a straight line 
and such that x'(?) / 0. Let 

e(?) = x(?) + -N(?). 

K 

This is the path traced by the center of the osculating circle of 
the path x. The quantity p = \/k is the radius of the osculat- 
ing circle and is called the radius of curvature of the path x. 
The path e is called the evolute of the path x. Exercises 20—25 
involve evolutes of paths. 

20. Letx(?) = (?, ? 2 )beaparabolicpath.(SeeFigure3.50.) 

(a) Find the unit tangent vector T, the unit normal 
vector N, and the curvature k as functions of ?. 

(b) Calculate the evolute of x. 



Miscellaneous Exercises for Chapter 3 241 



(c) Use a computer to plot x(f) and e(?) on the same 
set of axes. 



y 




Figure 3.50 The parabola and its 
osculating circle at a point. The centers 
of the osculating circles at all points of 
the parabola trace the evolute of the 
parabola as described in Exercise 20. 



21 . Show that the evolute of a circular path is a point. 

22. (a) Use a computer algebra system to calculate the for- 

mula for the evolute of the elliptical path x(f) = 
(a cost, b sinr). 

(b) Use a computer to plot x(f ) and the evolute e(f ) on 
the same set of axes for various values of the con- 
stants a and b. What happens to the evolute when 
a becomes close in value to bl 

23. Use a computer algebra system to calculate the formula 
for the evolute of the cycloid x(f ) = (at — a sin t, a — 
a cost). What do you find? 

24. Use a computer algebra system to calculate the formula 
for the evolute of the cardioid x(f) = (2a cosf(l + 
a cos t), 2a sin t (1 + a cos t )). 

25. Assuming ic'(t) ^ 0, show that the unit tangent vector 
to the evolute e(f ) is parallel to the unit normal vector 
N(f ) to the original path x(f). 

26. Suppose that a C 1 path x(f ) is such that both its veloc- 
ity and acceleration are unit vectors for all t . Show that 
k = 1 for all t. 

27. Consider the plane curve parametrized by 



x ( s ) = / cos g(t)dt, y(s) = I sin g(t)dt, 
Jo Jo 

where g is a differentiable function. 

(a) Show that the parameter s is the arclength param- 
eter. 

(b) Calculate the curvature k(s). 



(c) Use part (b) to explain how you can create a 
parametrized plane curve with any specified con- 
tinuous, nonnegative curvature function k(s). 

(d) Give a set of parametric equations for a curve 
whose curvature k(s) = \s\. (Your answer should 
involve integrals.) 

(e) Use a computer to graph the curve you found in 
part (d), known as a clothoid or a spiral of Cornu. 
(Note: The integrals involved are known as Fres- 
nel integrals and arise in the study of optics. You 
must evaluate these integrals numerically in order 
to graph the curve.) 

28. Suppose that x is a C 3 path in R 3 with torsion r always 
equal to 0. 

(a) Explain why x must have a constant binormal vec- 
tor (i.e., one whose direction must remain fixed for 
all f). 

(b) Suppose we have chosen coordinates so that x(0) = 
0 and that v(0) and a(0) lie in the .ty-plane (i.e., 
have no k-component). Then what must the binor- 
mal vector B be? 

(c) Using the coordinate assumptions in part (b), show 
that x(f) must lie in the jc;y-plane for all t . (Hint: 
Begin by explaining why v(f) • k = a(r) • k = Ofor 
all t. Then show that if 

x(t) = x(t)i + y(t)j + z(t)k, 

we must have z(t ) = 0 for all t .) 

(d) Now explain how we may conclude that curves 
with zero torsion must lie in a plane. 

29. Suppose that x is a C 3 path in R 3 , parametrized by arc- 
length, with k / 0. Suppose that the image of x lies in 
the xy -plane. 

(a) Explain why x must have a constant binormal 
vector. 

(b) Show that the torsion r must always be zero. 
Note that there is really nothing special about the im- 
age of x lying in the Jfj-plane, so that this exercise, 
combined with the results of Exercise 28, shows that 
the image of x is a plane curve if and only if r is always 
zero and if and only if B is a constant vector. 

30. In Example 7 of §3.2 we saw that if x is a straight-line 
path, then x has zero curvature. Demonstrate the con- 
verse; that is, if x is a C 2 path parametrized by arc- 
length s and has zero curvature for all s, then x traces 
a straight line. 

31 . A large piece of cylindrical metal pipe is to be manu- 
factured to include a strake, which is a spiraling strip 
of metal that offers structural support for the pipe. (See 
Figure 3.51.) The pieces of the strake are to be made 
from flat pieces of flexible metal whose curved sides 
are arcs of circles as shown in Figure 3.52. Assume that 



242 Chapter 3 | Vector- Valued Functions 



the pipe has a radius of a ft and that the strake makes 
one complete revolution around the pipe every h ft. 3 




Figure 3.51 A cylindrical Figure 3.52 A section of 
pipe with strake attached. the strake. (See Exercise 3 1 .) 

(a) In terms of a and h, what should the inner radius r 
be so that the strake will fit snugly against the pipe? 

(b) Suppose a = 3 ft and h = 25 ft. What is r? 

Suppose that x: / — > R 3 is a path of class C 3 parametrized by 
arclength. Then the unit tangent vector T(s) defines a vector- 
valued function T: / —> R 3 that may also be considered to be a 
path (although not necessarily one parametrized by arclength, 
nor necessarily one with nonvanishing velocity). Since T is a 
unit vector, the image of the path T must lie on a sphere of 
radius 1 centered at the origin. This image curve is called the 
tangent spherical image of x. Likewise, we may consider the 
functions defined by the normal and binormal vectors N and B 
to give paths called, respectively, the normal spherical image 
and binormal spherical image of x. Exercises 32-35 concern 
these notions. 

32. Find the tangent spherical image, normal spherical 
image, and binormal spherical image of the circular 
helix x(/) = (a cosf , a sinf , bt). (Note: The path x is 
not parametrized by arclength.) 

33. Suppose that x is parametrized by arclength. Show 
that x is a straight-line path if and only if its tangent 
spherical image is a constant path. (See Example 7 of 
§3.2 and Exercise 30.) 

34. Suppose that x is parametrized by arclength. Show that 
the image of x lies in a plane if and only if its binormal 
spherical image is constant. (See Exercises 28 and 29.) 

35. Suppose that x is parametrized by arclength. Show 
that the normal spherical image of x can never be 
constant. 

36. In this problem, we will find expressions for velocity 
and acceleration in cylindrical coordinates. We begin 



with the expression 

x(t) = x(t)i + y(t)j + z(t)k 

for the path in Cartesian coordinates. 

(a) Recall that the standard basis vectors for cylindri- 
cal coordinates are 

e,- = cosf? i + s'md j, 
ee = — s'md i + cos# j, 
e, = k. 

Use the facts that x = r cos 0 and y = r sin 0 to 
show that we may write x(r ) as 

x(t) = r(t)e r +z(t)e z . 

(b) Use the definitions of e r , ee, and e z just given and 
the chain rule to find de r /dt, deg/dt, and de./dt 
in terms of e r , eg , and e z . 

(c) Now use the product rule to give expressions for v 
and a in terms of the standard basis for cylindrical 
coordinates. 

37. Suppose that the path 

x(f) = (sin2f, Vlcoslt, sin2f — 2) 

describes the position of the Starship Inertia at time t . 

(a) Lt. Commander Agnes notices that the ship is trac- 
ing a closed loop. What is the length of this loop? 

(b) Ensign Egbert reports that the Inertia's path is 
actually a flow line of the Martian vector field 
F(jc, y, z) = yi — 2x\ + yk, but he omitted a con- 
stant factor when he entered this information in 
his log. Help him set things right by finding the 
correct vector field. 

38. Suppose that the temperature at points inside a room is 
given by a differentiable function T(x, y, z). Livinia, 
the housefly (who is recovering from a head cold), is in 
the room and desires to warm up as rapidly as possible. 

(a) Show that Livinia 's path x(?) must be a flow line 
of kVT, where & is a positive constant. 

(b) If T(x, y, z) = x 2 — 2y 2 + 3z 2 and Livinia is ini- 
tially at the point (2,3,-1), describe her path 
explicitly. 

39. Let F = u(x, y)i — v(x, y)j be an incompressible, 
irrotational vector field of class C 2 . 

(a) Show that the functions u and v (which deter- 
mine the component functions of F) satisfy the 
Cauchy-Riemann equations 

9m 3d du dv 

— = — , and — = . 

dx dv dy ox 



3 See F. Morgan, Riemannian Geometry: A Beginner's Guide, 2nd ed. (A K Peters, Wellesley, 1998), 
pp. 7-10. Figures 3.51 and 3.52 adapted with permission. 



Miscellaneous Exercises for Chapter 3 243 



(b) Show that u and v are harmonic, that is, that 



(Also see §1.4 concerning the notion of torque.) Show 
that 



d 2 u d 2 U 
9^ + 9^2 



40. Suppose that a particle of mass m travels along a path 
x according to Newton's second law F = ma, where 
F is a gradient vector field. If the particle is also con- 
strained to lie on an equipotential surface of F, show 
that then it must have constant speed. 

41 . Let a particle of mass m travel along a differentiable 
path x in a Newtonian vector field F (i.e., one that 
satisfies Newton's second law F = ma, where a is the 
acceleration of x). We define the angular momen- 
tum 1(f) of the particle to be the cross product of the 
position vector and the linear momentum mv, that is, 

l(r) = x(f) x m\(t). 

(Here v denotes the velocity of x.) The torque about 
the origin of the coordinate system due to the force F 
is the cross product of position and force: 

M(f ) = x(f) x F(f) = x(f) x ma(f). 



dt 



M. 



Thus, we see that the rate of change of angular mo- 
mentum is equal to the torque imparted to the particle 
by the vector field F. 

42. Consider the situation in Exercise 41 and suppose that 
F is a central force (i.e., a force that always points 
directly toward or away from the origin). Show that in 
this case the angular momentum is conserved, that is, 
that it must remain constant. 

43. Can the vector field 

F = (e x cos y + e~ x sin z) i — e x sin y j + e~ x cos z k 

be the gradient of a function f(x, y, z) of class C 2 ? 
Why or why not? 

44. Can the vector field 

F = *(y 2 + \)i + (ye x - e z ) j + x 2 e z k 

be the curl of another vector field G(x, y, z) of class 
C 2 ? Why or why not? 



4 Maxima and Minima 
in Several Variables 



4.1 Differentials and Taylor's 
Theorem 

4.2 Extrema of Functions 

4.3 Lagrange Multipliers 

4.4 Some Applications of 
Extrema 

True/False Exercises for 
Chapter 4 

Miscellaneous Exercises 
for Chapter 4 



4.1 Differentials and Taylor's Theorem 

Among all classes of functions of one or several variables, polynomials are without 
a doubt the nicest in that they are continuous and differentiable everywhere and 
display intricate and interesting behavior. Our goal in this section is to provide 
a means of approximating any scalar-valued function by a polynomial of given 
degree, known as the Taylor polynomial. Because of the relative ease with which 
one can calculate with them, Taylor polynomials are useful for work in computer 
graphics and computer-aided design, to name just two areas. 

Taylor's Theorem in One Variable: A Review 

Suppose you have a function /:lcR->R that is differentiable at a point a in 
X. Then the equation for the tangent line gives the best linear approximation for 
/ near a. That is, when we define p\ by 

p\{x) = f(a) + f'(a)(x — a), we have pi(x) & fix) if x % a. 

(See Figure 4.1.) As explained in §2.3, the phrase "best linear approximation" 
means that if we take R\(x, a) to be f(x) — p\(x), then 



lim 



Ri(x, a) 



0. 



Note that, in particular, we have pi(a) = f(a) and p[(a) = f'(a). 

Generally, tangent lines approximate graphs of functions only over very small 
neighborhoods containing the point of tangency. For a better approximation, we 
might try to fit a parabola that hugs the function's graph more closely as in 
Figure 4.2. In this case, we want p 2 to be the quadratic function such that 

p 2 (a) = f(a), p' 2 (a) = f'(a), and p' 2 \a) = f"(a). 

The only quadratic polynomial that satisfies these three conditions is 

P2 (x) = f(a) + f(a){x -a)+ ^(x - a) 2 . 



It can be proved that, if / is of class C 2 , then 



fix) = p 2 (x)+ R 2 (x, a), 



4.1 | Differentials and Taylor's Theorem 245 




where 




Figure 4.3 

fix) = In* 



y = 2- 

Approximations to 



Riix, a) 
lim = 0. 

*-»-a ix — a) 1 



EXAMPLE 1 If fix) = In x, then, for a = 1 , we have 

/(l) = lnl=0, 
1 
I 

1 

12 



/'(I) 
/"(I) = 



1. 



1. 



Hence, 



piix) = 0 + 1(* 
p 2 (x) = 0+ 1(jc 



1) = *- 1, 

1) - I(* - l) 2 = 



2x 



The approximating polynomials p\ and p2 are shown in Figure 4.3. ♦ 

There is no reason to stop with quadratic polynomials. Suppose we want to 
approximate / by a polynomial of degree k, where k is a positive integer. 
Analogous to the work above, we require that p k and its first k derivatives agree 
with / and its first k derivatives at the point a. Thus, we demand that 



Pk(a) = 


f{a), 


Pk( a ) = 


f'(a), 


P'l(a) = 





pf («) = / w (a). 

Given these requirements, we have only one choice for p k , stated in the following 
theorem: 



- f(k)( 



THEOREM 1.1 (Taylor's theorem in one variable) Let X be open in 
R and suppose /:XcR^R is differentiable up to (at least) order k. 



246 Chapter 4 I Maxima and Minima in Several Variables 



Given a e X, let 

f"(a) f( k Xa) 
p k (x) = f{a) + f'(a)(x -a)+ J -^-(x - af + • • • + J —^-{x - af. (1) 

2 k\ 

Then 

f(x) = Pk(x)+ Rk{x,a), 
where the remainder term R k is such that R k (x, a)/(x — af -¥ 0 as x —* a. 




Figure 4.4 The graphs of 

(1) y =x -a, 

(2) y = (x — a) 2 , and 

(3) y = (x- af. 

Note how much more closely the 
graph of (3) hugs the jc-axis than 
thatof(l)or(2). 



The polynomial defined by formula (1) is called the &th-order Taylor poly- 
nomial of / at a. The essence of Taylor's theorem is this: For x near a, the Taylor 
polynomial p k approximates / in the sense that the error R k involved in making 
this approximation tends to zero even faster than (x — af does. When k is large, 
this is very fast indeed, as we see graphically in Figure 4.4. 

EXAMPLE 2 Consider In x with a = 1 again. We calculate 
/(l) = lnl = 0, 

/'(l) = y = l> 



/"(I) 



1 

12 



1* 



i^l = (-l) k+ \k - 1)!. 



Therefore, 

Pk(x) = (x 



lf + - 3 (x 



If 



+ 



(-1) 



<X ~ I f 



Taylor's theorem as stated in Theorem 1.1 says nothing explicit about the 
remainder term R k . However, it is possible to establish the following derivative 
form for the remainder: 



PROPOSITION 1 .2 If / is of class C k+ \ then there exists some number z be- 
tween a and x such that 



R k (x, a) = f( + )(Z \ x - af + \ 
(Jfc + 1)! 



(2) 



In practice, formula (2) is quite useful for estimating the error involved with a 
Taylor polynomial approximation. Both Theorem 1 . 1 (under the slightly stronger 
hypothesis that / is of class C k+ 1 ) and Proposition 1 .2 are proved in the addendum 
to this section. 



EXAMPLE 3 The fifth-order Taylor polynomial of f(x)= cos x about x = tt/2 
is i « 1 < 

/ 7T\ 1 / 7T\ 3 1 / 7T\ 5 

P5(x) = -\ x -2) + 6\ x -2) -mV'V ■ 



4.1 | Differentials and Taylor's Theorem 247 



(You should verify this calculation.) According to formula (2), the difference 
between p 5 and cos x is 



R, 



( n \ f (6 \z) ( 11 \ 6 COSZ / 7t\ ( 

\ X '2) = ~^\ X ~2) = ~~6T\ X ~2) 



2/ 6! V 2/ 6! V 2. 

where z is some number between n/2 and x. Since | cosjcI is never larger than 1, 
we have 

^ 7T\6 ^(jC-TT/2) 6 



*5 (*, |) 



cos z 



6! 



Thus, for x in the interval [0, it], we have 

(tt - 7T/2) 6 



R 5 (x. |) 



71' 



720 



46,080 



720 



0.0209. 



In other words, the use of the polynomial ps above in place of cosx will be 
accurate to at least 0.0209 throughout the interval [0, n]. ♦ 

Taylor's Theorem in Several Variables: 

The First-order Formula 



For the moment, suppose that /: X c R 2 — > R is a function of two variables, 
where X is open in R 2 and of class C l . Then near the point (a, b) e X, the best 
linear approximation to / is provided by the equation giving the tangent plane at 
(a, b, f(a, b)). That is, 

f(x, y) » pi(x, v), 

where 

pi(x, y) = f(a, b) + f x (a, b)(x -a) + f y (a, b)(y - b). 

Note that the linear polynomial p\ has the property that 

pi(a, b) = f(a, b); 

dpi df 

-7— {a,b)= —(a,b), 

ox ox 

dpi df 
-5— (a, b) = —(a, b). 
oy oy 

Such an approximation is shown in Figure 4.5. 

To generalize this situation to the case of a function /: X c R" — > R of 
class C 1 , we naturally use the equation for the tangent hyperplane. That is, if 



z=f(x,y) 




Figure 4.5 The graph of z = f(x, y) and 
z = pi(x, y). 



248 Chapter 4 I Maxima and Minima in Several Variables 



a = (ai, a 2 , . . . , a„) e X, then 

f(Xl,X2, ...,x n )& pi{xi,x 2 , ...,x„), 

where 

Pi(xi, x n ) = /(a) + f xi (a)(xi - ai) + / X2 (a)(x 2 - a 2 ) 

+ r- fx„( a )( x n ~ a n)- 

Of course, the formula for p\ can be written more compactly using either 
E -notation or, better still, matrices: 

n 

Pi(jci, . . . , x n ) = /(a) + £ /*(»)(* - a,) = /(a) + D/(a)(x - a). (3) 

; = 1 

EXAMPLE 4 Let /(xj, X2, X3, X4) = xi + 2x 2 + 3x3 + 4x4 + X1X2X3X4. Then 

3/ 1 3/ „ 

= l + X2X3X4, = 2 + X1X3X4, 

3xj 3x2 

3/ „ 3/ 

- — = 3+xix 2 x 4 , - — =4 + X!X 2 x 3 . 

dX3 dX4 

At a = 0 = (0, 0, 0, 0), we have 

^(0) = l, ^(0) = 2, |^(0) = 3, |^(0) = 4. 

dx\ 6x2 0x3 dx4 

Thus, 

piixi, x 2 , x 3 , x 4 ) = 0 + l(xi - 0) + 2(x 2 - 0) + 3(x 3 - 0) + 4(x 4 - 0) 

= X\ + 2X2 + 3X3 + 4X4. 

Note that p\ contains precisely the linear terms of the original function /. On the 
other hand, if a = (1 , 2, 3, 4), then 

-Ml, 2, 3, 4) = 25, —^-(1,2, 3,4) = 14, 
dxi dx2 

df df 

-Ml, 2, 3, 4)= 11, -Ml, 2, 3, 4) = 10, 

dX3 dX4 

so that, in this case, 

pi(xi,X2,x 3 ,x 4 ) = 54 + 25(xi - 1)+ 14(x 2 - 2) + ll(x 3 -3)+ 10(x 4 - 4). 

♦ 

The relevant theorem regarding the first-order Taylor polynomial is just a re- 
statement of the definition of differentiability. However, since we plan to consider 
higher-order Taylor polynomials, we state the theorem explicitly. 

THEOREM 1 .3 (First-order Taylor's formula in several variables) Let 
X be open in R" and suppose that /:XcR"->R is differentiate at the point 
a in X. Let 

pi(x) = /(a) + D/(aXx-a)- (4) 

Then 

/(x) = pi(x) + *i(x, a), 
where Ri(x, a)/||x — a|| — >• 0 as x —> a. 



4.1 | Differentials and Taylor's Theorem 249 

Note that we may also express the first-order Taylor polynomial using the gradient. 
In place of (4), we would have 

Pi(x) = /(a) + V/(a).(x-a). 

Differentials 

Before we explore higher-order versions of Taylor's theorem in several variables, 
we consider the linear (or first-order) approximation in further detail. 
Let h = x — a. Then formula (3) becomes 

Pl (x) = /(a) + fl/(a)h = m + ■ (5) 

i=i dx > 

We focus on the sum appearing in formula (5) and summarize its salient 
features as follows: 



DEFINITION 1 .4 Let / : X C R" -> R and let ael The incremental 
change of /, denoted Af, is 

A/ = /(a + h)-/(a). 

The total differential of /, denoted df(a, h), is 

df df df 

dX\ 0X2 dx n 

The significance of the differential is that for h « 0, 

Af « df. 
(We have abbreviated df(a, h) by J /.) 



Sometimes hj is replaced by the expression Ax, or dx { to emphasize that it 
represents a change in the z'th independent variable, in which case we write 

At 9/ A A V A 

df = —dx l + —dx 2 -\ h — dx n . 

dx\ 0X2 6x n 

(We've suppressed the evaluation of the partial derivatives at a, as is customary.) 
EXAMPLE 5 Suppose f(x, y, z) = sin(xyz) + cos(xyz). Then 

Af V j . &f , df 
df = — dx + — dy + — dz 
6x 6y 6z 

= yz[cos(xyz) — sm(xyz)]dx + xz[cos(xyz) — sin(xyz)]dy 

+ xy[cos(xvz) — sin(xyz)]dz 
= (cos(xyz) — sin(xyz))(yz dx + xz dy + xy dz)- ^ 

The geometry of the differential arises, naturally enough, from tangent lines 
and planes. (See Figures 4.6 and 4.7.) In particular, the incremental change Af 
measures the change in the height of the graph of / when moving from a to a + h; 
the differential change d f measures the corresponding change in the height of 



250 Chapter 4 [ Maxima and Minima in Several Variables 




Figure 4.6 The incremental 
change A/ equals the change in 
y -coordinate of the graph of 
y = f(x) as the x-coordinate of a 
point changes from a to a + dx. 
The differential df equals the 
change in y -coordinate of the 
graph of the tangent line at a (i.e., 
the graph of y = p\(x)). 



Figure 4.7 The incremental 
change Af equals the change in 
^-coordinate of the graph of 
z = f(x, y) as a point in R 2 
changes from a = (a, b) to 
a + h = (a + h, b + k). The 
differential df equals the change in 
z-coordinate of the graph of the 
tangent plane at (a, b). 



the graph of the (hyper)plane tangent to the graph at a. When h is small (i.e., 
when a + h is close to a), the differential df approximates the increment Af and 
it is often easier from a technical standpoint to work with the differential. 

EXAMPLE 6 Let /(*, y) = x - y + 2x 2 + xy 2 . Then for (a, b) = (2,-1), 
we have that the increment is 

Af = f(2 + Ax, -1 + A.y) - f(2, -1) 

= 2 + Ax - (-1 + Ay) + 2(2 + Ax) 2 + (2 + Ax)(-1 + A.y) 2 - 13 
= lOAx - 5 Ay + 2(Ax) 2 - 2Ax Ay + 2(A^) 2 + Ax(Ay) 2 . 

On the other hand, 

df((2, -1), (Ax, Ay)) = f x (2, -I) Ax + f y (2, -I) Ay 

= (1 + 4x + .v 2 )| (2 ,-i)Ax + (-1 + 2x>>)|(2,-i)A.y 
= lOAx - 5A;y. 

We see that df consists of exactly the terms of Af that are linear in Ax and Ay 
(i.e., appear to first power only). This will always be the case, of course, since 
that is the nature of the first-order Taylor approximation. Use of the differential 
approximation is often sufficient in practice, for when Ax and Ay are small, higher 
powers of them will be small enough to make virtually negligible contributions 
to Af. For example, if Ax and Ay are both 0.01, then 

df = (0.1 -0.05) = 0.05 

and 

Af = (0.1 - 0.05) + 0.0002 - 0.0002 + 0.0002 + 0.000001 
= 0.05 + 0.000201 = 0.050201. 

Thus, the values of df and Af are the same to three decimal places. ♦ 



4.1 | Differentials and Taylor's Theorem 251 



EXAMPLE 7 A wooden rectangular block is to be manufactured with dimen- 
sions 3 in x 4 in x 6 in. Suppose that the possible errors in measuring each di- 
mension of the block are the same. We use differentials to estimate how accurately 
we must measure the dimensions so that the resulting calculated error in volume 
is no more than 0.1 in 3 . 

Let the dimensions of the block be denoted by x 3 in), y 4 in), and z 
6 in). Then the volume of the block is 

V=xyz and V « 3 ■ 4 • 6 = 72 in 3 . 

The error in calculated volume is A V, which is approximated by the total differ- 
ential dV. Thus, 

A V w d V = V x (3, 4, 6)Ax + V y (3, 4, 6)Ay + V,(3, 4, 6)A Z 
= 24Ax + 18Ay + 12Az. 

If the error in measuring each dimension is e, then we have Ax = Ay = Az = e. 
Therefore, 

dV = 24Ax + 18Ay + 12Az = 24e + 18e + 12e = 54e. 
To ensure (approximately) that | A V\ < 0.1, we demand 

\dV\ = |54e| < 0.1. 

Hence, 

0.1 

|e| < — = 0.0019 m. 
" 54 

So the measurements in each dimension must be accurate to within 0.0019 in. ♦ 





Figure 4.8 Which would 
you buy? 



EXAMPLE 8 The formula for the volume of a cylinder of radius r and height 
h is V(r, h) = jtr 2 h. If the dimensions are changed by small amounts Ar and 
Ah, then the resulting change A V in volume is approximated by the differential 
change d V. That is, 



AV ^dV = 



dV dV , 

— Ar H Ah = InrhAr + nr Ah. 

dr dh 



Suppose the cylinder is actually a beer can, so that it has approximate dimensions 
of r = 1 in and h = 5 in. Then 

dV = jr(10Ar + Ah). 

This statement shows that, for these particular values of r and h, the volume 
is approximately 10 times more sensitive to changes in radius than changes in 
height. That is, if the radius is changed by an amount e, then the height must be 
changed by roughly lOe to keep the volume constant (i.e., to make AV zero). 
We use the word "approximate" because our analysis arises from considering the 
differential change d V rather than the actual incremental change A V. 

This beer can example has real application to product marketing strategies. 
Because the volume is so much more sensitive to changes in radius than height, 
it is possible to make a can appear to be larger than standard by decreasing its 
radius slightly (little enough so as to be hardly noticeable) and increasing the 
height so no change in volume results. (See Figure 4.8.) This sensitivity analysis 
shows that even a tiny decrease in radius can force an appreciable compensating 
increase in height. The result can be quite striking, and these ideas apparently 



252 Chapter 4 I Maxima and Minima in Several Variables 



have been adopted by at least one brewery. Indeed this is how the author came to 
fully appreciate differentials and sensitivity analysis. 1 ♦ 

Taylor's Theorem in Several Variables: 

The Second-order Formula 

Suppose /: X c R 2 —> R is a C 2 function of two variables. Then we know that 
the tangent plane gives rise to a linear approximation p\ of / near a given point 
(a, b) of X. We can improve on this result by looking for the quadric surface that 
best approximates the graph of z = fix, y) near (a, b, f(a, b)). See Figure 4.9 
for an illustration. That is, we search for a degree 2 polynomial pi(x, y) = Ax 2 + 
Bxy + Cy 2 + Dx + Ey + F such that, for (x, y) (a, b), 

f(x, y) « p 2 (x, y). 



Quadric 
surface 




Z=f(x,y) 



Figure 4.9 The tangent plane and quadric 
surface. 



Analogous to the linear approximation pi, it is reasonable to require that p2 and 
all of its first- and second-order partial derivatives agree with those of / at the 
point (a, b). That is, we demand 



dpi 
dx 

d 2 P2 
dx 2 



(a, b) = 



(a,b) 



p 2 (a, b) 

— ( 

dx 

d 2 f 
— — ( 
dx 2 

d 2 p 2 



(a,b), 
(a, b), 

2 (a, b) 



f(a,b) 
dp 2 



dy 

d 2 P2 
dxdy 

d 2 f 



(a, b) = 



(a,b) = 



df 

dy 
d 2 f 



dxdy 



{a, b), 
(a,b), 



(6) 



dy 



dy : 



(a,b). 



After some algebra, we see that the only second-degree polynomial meeting these 
requirements is 

P2(x, y) = f(a, b) + f x (a, b)(x -a) + f y (a, b)(y - b) 



+ y xx (a, b)(x - a) 2 + f xy (a, b)(x - a)(y - b) 
+ \f yy (a,b)(y-b) 2 . 



(7) 



1 See S. J. Colley, The College Mathematics Journal, 25 (1994), no. 3, 226-227. Art reproduced with 
permission from the Mathematical Association of America. 



4.1 | Differentials and Taylor's Theorem 253 



How does formula (7) generalize to functions of n variables? We need to begin 
by demanding conditions analogous to those in (6) for a function /: X cR"->R. 
For a = (a\, a 2 , . . . , a n ) £ X, these conditions are 

p 2 (a) = /(a), 

-^(a)=-^(a), i = l,2,...,n, (8) 

dxi dxi 

d 2 p 2 d 2 f 

(a)= — ^-(a), ij = 1,2,..., n. 



dxidxj dxidxj 

If you do some algebra (which we omit), you will find that the only polynomial 
of degree 2 that satisfies the conditions in (8) is 

n \ H 

P2(x) = /(a) + A,( a )(*< " a d +2 E fxi*M( X i ~ a i)( x J - ( 9 ) 

1=1 i.j=l 

(Note that the second sum appearing in (9) is a double sum consisting of n 2 terms.) 
To check that everything is consistent when n = 2, we have 

p 2 (xi,x 2 ) = f{a\,a 2 ) + fxM^' a i)(xi - fli) + fx 2 (ai,a 2 )(x2 - a 2 ) 

+ \ \_Sx^xSfl\^i){x\ - aif + f XlX2 (ai, a 2 ){x\ - a { )(x 2 - a 2 ) 
+ fx 2Xl (ai,a 2 )(x 2 - a 2 ){x\ - d\) + f XlX2 {a\, a 2 )(x 2 - a 2 ) 2 ] . 

When /' is a C 2 function, the two mixed partials are the same, so this formula 
agrees with formula (7). 

EXAMPLE 9 Let /(*, y, z) = e x+y+z and let a = (a, b, c) = (0, 0, 0). Then 
/(0,0, 0) = e°= 1, 

f x (0, 0, 0) = /,(0, 0, 0) = f z (0, 0, 0) = e° = 1, 
f xx (Q, 0, 0) = f xy (0, 0, 0) = f xt (0, 0, 0) = f yy (0, 0, 0) 
= f yz (0, 0, 0) = /„((), 0, 0) = e° = 1. 

Thus, 

P2 (x, y, z ) = l + l(x - 0) + l(y - 0) + l(z - 0) 

+ \ [l(x - 0) 2 + 2 ■ l(x - 0)(y - 0) + 2 ■ l(x - 0)(z - 0) 
+ l(y - 0) 2 + 2 • l(y - 0)(z - 0) + l(z - 0) 2 ] 

= 1 + x + y + z + k 2 + xy + xz + + yz + \z 2 



= 1 + (x + y + z) + Ux + y + zf 



2 



We have made use of the fact that, since / is of class C , a term like 

f xy (0, 0, 0)(x - 0)(y - 0) is equal to f yx (0, 0, 0)(y - 0)(x - 0). 



Now we state the second-order version of Taylor's theorem precisely. 



254 Chapter 4 I Maxima and Minima in Several Variables 



THEOREM 1 .5 (Second-order Taylor's formula) Let X be open in R" , and 
suppose that /:XcR"->R is of class C 2 . Let 

n j n 

P2OO = /(a) + fxi( & X x i ~ a d + j Y fxixMXXi ~ a ^ X J ~ a i)" 
i=l Z <,J=1 

Then 

/(x) = p 2 (x) + fl 2 (x, a), 
where |/? 2 |/||x - all 2 -> 0 as x ->• a. 

A version of Theorem 1.5, under the stronger assumption that / is of class C 3 , is 
established in the addendum to this section. 

EXAMPLE 10 Let f(x, y) = cosx cosy and (a, b) = (0, 0). Then 

/(0,0) = 1; 

f x (0, 0) = - sin x cosy I (0,0) = 0, f y (0, 0) = - cosx siny| (0 , 0 ) = 0; 
f xx (0, 0) = - cosxcosy| (0 0) = -1, 

f xy (Q, 0) = sinx siny| (0 0) = 0, 

/y y (0,0) = - cosxcosy| (00) = -1. 

Hence, 

f(x, y) « p 2 (x, y) = 1 + \{-\ ■ x 2 - 1 • y 2 ) = 1 - \x 2 - \y 2 . 

We can also solve this problem another way since / is a product of two functions. 
We can multiply the two Taylor polynomials: 

Pi(x, y) = (Taylor polynomial for cos x) • (Taylor polynomial for cos y) 
= (1 - \x 2 ) (1 - \y 2 ) 

= 1 — \x 2 — \y 2 up to terms of degree 2. 

This method is justified by noting that if qi is the Taylor polynomial for cosine 
and Ro is the corresponding remainder term, then 

cosx cosy = [q 2 (x) + R 2 (x, 0)][q 2 (y) + R 2 (y, 0)] 

= qi(x)q 2 (y) + q 2 (y)R 2 (x, 0) + q 2 (x)R 2 (y, 0) + R 2 (x, 0)R 2 (y, 0) 

= q 2 (x)q 2 (y) + other stuff, 

where (other stuff)/||(x, y)|| 2 —> 0 as (x, y) (0, 0), since both R 2 {x, 0) and 
R 2 (y, 0) do. ♦ 



The Hessian 

Recall that the formula for the first-order Taylor polynomial p\ was written quite 
concisely in formula (5) by using vector and matrix notation. It turns out that it 
is possible to do something similar for the second-order polynomial p 2 . 



4.1 | Differentials and Taylor's Theorem 255 



DEFINITION 1 .6 The Hessian of a function /:XcR"->R is the matrix 


\x/ , Vir\cf 1 / t 1"Vi ptitr\7 ic f 1 r) v 
YVllUov £ / 111 C11L1 y la u / / (JA 


j OJij . 1 Hal la, 

fx,X! fx,X 2 


fx\x„ 




Hf = 


fx 2 X\ fx 2 x 2 


fx 2 x„ 




_ fx„xi fx„x 2 


fx n x„ 





The term "Hessian" comes from Ludwig Otto Hesse, the mathematician who 
first introduced it, not from the German mercenaries who fought in the American 
revolution. 

Now let's look again at the formula for p 2 in Theorem 1.5: 

n j n 

P2 (x) = /(a) + £ f Xi (*)hi + fx,* (»)/<;/',. 

(We have let h = (hi, . . . , h n ) = x — a.) This can be written as 



p 2 « = /(a)+[/, 1 (a) / T2 (a) ■■■ A„(a)] 



&2 



+ 



1 r 



/i 2 ■■• h„ 



Thus, we see that 



/* 2 »(a) A 2 * 2 (a) ••• A 2X „(a) 
_/*„*, (a) /x„.v 2 (a) ••• A,,,, (a) J _ 



h 2 



p 2 (x) = /(a) + D/(a)h + ^h r ///(a)h. (10) 



(Remember that h T is the transpose of the w x 1 matrix h.) 

EXAMPLE 11 (Example 10 revisited) For f(x, y) = cosx cos y, a = (0, 0), 

we have 

Df(x, y) = [— sinx cosy — cosx sin y] 

and 



Hf(x,y) 



cos x cos y sin x sin y 
sin x sin y — cos x cos y 



256 Chapter 4 [ Maxima and Minima in Several Variables 



Hence, 



P2 (x, y) = /(O, 0) + Df(0, 0)h + lh T Hf(0, 0)h 



l+[ 0 0] 



+ \[h, h 2 ] 



' -1 0 " 




" hi ' 


0 -1 







= 1 



l -h 2 

2 1 



X -h 2 
2 n 2- 



Once we recall that h = (hi, h 2 ) = (x — 0, y — 0) = (x, y), we see that this result 
checks with our work in Example 10, just as it should. ♦ 



Higher-order Taylor Polynomials 

So far we have said nothing about Taylor polynomials of degree greater than 2 
in the case of functions of several variables. The main reasons for this are (i) the 
general formula is quite complicated and has no compact matrix reformulation 
analogous to (10) and (ii) we will have little need for such formulas in this text. 
Nonetheless, if your curiosity cannot be denied here is the third-order Taylor 
polynomial for a function /: X c R" —> R of class C 3 near aeX: 



P3(x) = /(a) + fxi(*)(xi ~ ad + ~ ^ - atX x j - aj) 

i=l i,j=\ 

1 " 

i,j,k=l 



(The relevant theorem regarding pi is that f(x) — pj,(x) + Rj,(x, a), where 
|/?3(x, a)|/||x — a || 3 -> 0 as x -> a.) If you must know even more, the /cth-order 
Taylor polynomial is 



i=l Z i,y=l 

1 " 

H h jti XI A ( a ^ Xi 'i ~ fl, 'i) ' ' ' ^ ~ a <* )• 

l'l,...,H = l 



Formulas for Remainder Terms (optional) 

Under slightly stricter hypotheses than those appearing in Theorems 1.3 and 
1.5, integral formulas for the remainder terms may be derived as follows. Set 
h = x — a. If / is of class C 2 , then 

n p 1 

Ri(x, a) = V / (1 - / )/, ,, (a + th)h,hj dt 

= [ [h T #/(a + fh)h](l -t)dt. 
Jo 



4.1 | Differentials and Taylor's Theorem 257 



If / is of class C 3 , then 

v r<i 2 



^ (l - o 2 

R 2 (x, a) = 2_, / — ^ — ./',,«,, (a + th)hihjh k dt, 



and if / is of class C* +1 , then 

i?,(x,a)= J] / — — — fx iy x i2 -x it+1 (a + t\L)h h hi 2 ■ ■ ■ h h+l dt. 
U '*+i=i Jo 

Although explicit, these formulas are not very useful in practice. By artful appli- 
cation of Taylor's formula for a single variable, we can arrive at derivative versions 
of these remainder terms (known as Lagrange's form of the remainder) that are 
similar to those in the one-variable case. 



Lagrange's form of the remainder. If / is of class C 2 , then in Theorem 1 .3 
the remainder Ri is 



1 " 

#i(x, a) = - J] f XiXj (z)hihj 



2 UM 

for a suitable point z in the domain of / on the line segment joining a and 
x = a + h. Similarly, if / is of class C 3 , then the remainder R2 in Theorem 
1.5 is 



1 - 

tf 2 (x, a)= — f XiXjXk {z)hihjh k 



3 1 

ij,k=l 

for a suitable point z on the line segment joining a and x = a + h. More 
generally, if / is of class C k+l , then the remainder R k is 

*^ X ' a) = TkTiv fw-*** V)KK ■ ■ ■ K + i 

K <i,-,i*+i =1 

for a suitable point z on the line segment joining a and x = a + h. 



The remainder formulas above are established in the addendum to this section. 

EXAMPLE 12 For fix, y) = cos x cos y, we have 

2 

X] fx iXjXk (z)hihjh k 



\R 2 (x,y,0, 0)| = i 



2 



< 

tj,*=i 

since all partial derivatives of / will be a product of sines and cosines and, hence, 
no larger than 1 in magnitude. Expanding the sum, we get 

\R 2 (x,y,0, 0)| < i(|/ Jl | 3 + 3/i?|/! 2 | + 3|/i 1 |^ + |/z 2 | 3 ). 

If both \hi \ and \ fi2\ are no more than, say, 0.1, then 

\R 2 (x,y,0, 0)| < i (8- (O.l) 3 ) = 0.0013. 



Chapter 4 i Maxima and Minima in Several Variables 



y 



z 



0.1 




Z = cos x cos y 



-0.1 



0.1 



X 



y 



-0.1 



X 



Figure 4.11 The graph of f(x, y) = 
cos x cos y and its Taylor polynomial 



Figure 4.10 The polynomial p2 
approximates / to within 0.0013 
on the square shown. (See 
Example 12.) 



Pi(x, y) = 1 — \x 2 — \y 2 over the square 
{(Jt.y) 1-1 <*<!,-! <?<!}. 



So throughout the square of side 0.2 centered at the origin and shown in Fig- 
ure 4.10, the second-order Taylor polynomial is accurate to at least 0.0013 (i.e., to 
two decimal places) as an approximation of f(x, y) = cosx cosy. In Figure 4. 11, 
we show the graph of f(x, y) — cos x cos y over the square domain {(x , y) | — 1 < 
x < 1 , — 1 < y < 1} together with the graph of its second-order Taylor polyno- 
mial p2(x, y) — 1 — \x 2 — \y 2 (calculated in Example 10). Note how closely the 
surfaces coincide near the point (0, 0, 1), just as the analysis above indicates. ♦ 

Addendum: Proofs of Theorem 1.1, Proposition 1.2, 
and Theorem 1.5 

Below we establish some of the fundamental results used in this section. We begin 
by proving Theorem 1.1, Taylor's theorem for function of a single variable, and 
Proposition 1 .2 regarding the remainder term in Theorem 1.1. We then use these 
results to "bootstrap" a proof of the multivariable result of Theorem 1.5 and to 
derive Lagrange's formula for the remainder term appearing in it. 

Proof of Theorem 1 .1 We prove the result under the stronger assumption that / 
is of class C k+l rather than assuming that / is only differentiate up to order k. 
(This distinction matters little in practice.) 
By the fundamental theorem of calculus, 



We evaluate the integral on the right side of ( 1 1 ) by means of integration by parts. 
Recall that the relevant formula is 



We use this formula with u = f'(t) and v = x — t so that dv = —dt. (Note that 
in the right side of (1 1), x plays the role of a constant.) We obtain 




(11) 



u dv = uv — 





(12) 



4.1 | Differentials and Taylor's Theorem 259 



Combining (11) and (12), we have 

fix) = f(a) + f'(a)(x -a) + \\x - t)f"(t)dt. (13) 
Thus, we have shown, when / is differentiable up to (at least) second order, that 
Ri(x,a)= f (x - t)f"(t)dt. 

J a 

This provides an integral formula for the remainder in formula ( 1 ) of Theorem 1 . 1 
when k = 1, although we have not yet established that R\(x,a)/(x — a) — > 0 as 
x — > a. 

To obtain the second-order formula, the case k = 2 of (1), we focus on 
Ri(x, a) = f*(x — t)f"{t)dt and integrate by parts again, this time with u = 
f"(t) and v = (x - tf/2, so that dv = -(x - t)dt. We obtain 



f\x 

J a 



t)f"(t)dt = 



f"(t)(x - tf 



f"(a)(x - a) 2 



a Ja 

f 



f"'(t)dt 



(x - tf 



f"'(t)dt. 



Hence (13) becomes 

f"(a) 

f(x) = f(a) + f'(a)(x -a)+ J —^-{x 



2 r* (x-t) 

a) + L — 



Therefore, we have shown, when / is differentiable up to (at least) third order, 
that 

-Jlf"\t)dt. 



R 2 (x,a)= f 

J a 



We can continue to argue in this manner or use mathematical induction to show 
that formula (1) holds in general with 

R k (x,a) = j"^t^\t)dt, (14) 

assuming that / is differentiable up to order (at least) k+\. 

It remains to see that Rk(x, a)/(x — a) k — > 0 as x — > a. In formula (14) we 
are only considering t between a and x, so that \x — t \ < \x — a\. Moreover, since 
we are assuming that / is of class C k+1 , we have that f (k+x \t) is continuous and 
therefore, bounded for t between a and x (i.e., that \f {k+l \t)\ < M for some 
constant M). Thus, 



\R k (x,a)\ < 



r (x - tf 

Ja k\ 



f (k+ \t)dt 



< ± 



f 



(x - tf 



k\ 



f ik+l) (t) 



dt, 



where the plus sign applies if x > a and the negative sign if x < a, 



C x M 



x - a\ k dt 



M 



\k+\ 



Thus, 



Rk(x, a) 



(x — a) k 



M 



as x —> a, as desired. 



260 Chapter 4 I Maxima and Minima in Several Variables 



Proof of Proposition 1.2 We establish Proposition 1.2 by means of a general 
version of the mean value theorem for integrals. This theorem states that for 
continuous functions g and h such that h does not change sign on [a, b] (i.e., 
either h(t) > 0 on [a, b] or h(t) < 0 on [a, b]), there is some number z between 
a and b such that 

f g(t)h(t)dt = g(z) f h(t)dt. 

(We omit the proof but remark that this theorem is a consequence of the interme- 
diate value theorem.) Applying this result to formula (14) with g(t) = f (k+ ^(t) 
and h(t) = (x — t) k /k\, we find that there must exist some z between a and x 
such that 



i) = f (k+ \z)\ X{ ^-^dt=f^\z) 



k\ 



(x - t) 



(k+iy. 



(k + iy. 



(x — a) 



k+l 



Proof of Theorem 1.5 As in the proof of Theorem 1 . 1 , we establish Theorem 1 .5 
under the stronger assumption that / is of class C 3 . Begin by setting h = x — a, 
so that x = a + h, and consider a and h to be fixed. We define the one-variable 
function F by F(t) = /(a + th). Since / is assumed to be of class C 3 on an 
open set X, if we take x sufficiently close to a, then F is of class C 3 on an open 
interval containing [0, 1]. Thus, Theorem 1.1 with k = 2, a = 0, and x = 1 may 
be applied to give 



F(l) = F(0) + F'(0)(1 - 0) + ^9(1 - 0) 2 + R 2 (l, 0) 
F"(0) 

= F(0) + F'(0) + — ^ + R 2 (l, 0), 



(15) 



where R 2 (l, 0) = /J ^-F"'(0^.Now we use the chain rule to calculate deriva- 
tives of F in terms of partial derivatives of /: 



.7 = 1 



hi = ^ f XiXj (a + th)hihj; 



F'(t) = D/(a + th)h = J2 U ■(» + fh )^ 
F"(r) = £ 
F"'(t) = 

k=\ 

Thus, (15) becomes 

n j n 

/(a + h) = /(a) + fx,Wi + ~ E UM) h i h J 



J2 /u,,, ; (a • /h)/;,/;, 
.W=i 



fyfc = X! fxiXjxM + t^hihjhk 
i,j,k=\ 



+ 



n 1 



(l - o 2 



f XiXjXk (a + th)hihjh k dt, 



4.1 | Differentials and Taylor's Theorem 261 



or, equivalently, 

n \ n 

/(x) = /(a) + /*,(»)(*; " ad +~Y1 fxitjW&i ~ a i)( x i ~ a i) 

+ # 2 (x, a), 
where the multivariable remainder is 

R 2 (x,a)= V / y —— L f XiX]Xl (a + th)h i hjh k dt. 

>,j,k=\ Jo l 



(16) 



We must still show that |i?2(x, a)|/||x — a|| 2 -> 0 as x -> a, or, equivalently, 
that |^?2(x, a)|/|h|| 2 -> 0 as h -> 0. To demonstrate this, note that, for a and h 
fixed the expression (1 — t) 2 f XjXjXk (st + fh) is continuous for f in [0, 1] (since / 
is assumed to be of class C 3 ), hence bounded. In addition, for i = 1, . . . , n, we 



have that |/z,-| < ||h||. 


Hence, 


|tf 2 (x,a)| = 




< 


;i 1 


< 


n «1 



(i - o 2 

2 

(1 " tf 



fxtXjxM + th)hihjh k dt 
f XiXjXt (a + th)hjhjh k dt 
M||/i|| 3 dt = n 3 M||h|| 3 = n 3 M||x - a| 



Thus, 



|i? 2 (x, a)| , 

v '1 <« 3 M||x-a| 



x- a 



0 



as x -> a. 

Finally, we remark that entirely similar arguments may be given to establish 
results for Taylor polynomials of orders higher than two. ■ 

Lagrange's formula for the remainder (see page 257) Using the function 
F(t) = /(a + fh) defined in the proof of Theorem 1.5, Proposition 1.2 implies 
that there must be some number c between 0 and 1 such that the one-variable 
remainder is 

F"'(c) - 
tf 2 (l,0)=^-Al-0) 3 . 

Now, the remainder term Rz(I, 0) from Proposition 1.2 is precisely ^(x, a) in 
Theorem 1.5 and 

n n 

F"'(c)= f XiXiXt {* + c\v)hihjh k = f XiXjXk (z)hihjh k , 

i,j,k=\ i,j,k=l 

where z = a + ch. Since c is between 0 and 1 , the point z lies on the line segment 
joining a and x = a + h, and so 

1 



^2(x, a)=— f XiXjXk (z)hihjh k 



i,j,k=l 



which is the result we desire. The derivation of the formula for R k (x, a) for k > 2 
is analogous. ■ 



262 Chapter 4 I Maxima and Minima in Several Variables 

4.1 Exercises 



In Exercises 1—7, find the Taylor polynomials of given order 
k at the indicated point a. 

1. f( x ) = e 2x ,a = 0,k = A 

2. fix) = ln(l + x), a = 0,k = 3 

3. f(x)= l/x 2 ,a = l,k = 4 

4. f(x) = Jx,a = l,k = 3 

5. f(x) = </x, a = 9, k = 3 

6. f(x) = sinx, a = 0, k = 5 

7. f(x) = sinx, a = n/2, k = 5 

In Exercises 8— 15, find the first- and second-order Taylor poly- 
nomials for the given function f at the given poin t a. 

8. /(x,y) = l/(x 2 + y 2 + l),a = (0,0) 

9. f(x, y) = l/(x 2 + y 2 + 1), a = (1, -1) 

10. f(x,y) = e 2x+ y,a = (0, 0) 

11. f(x, y) = e 2x cos3y, a = (0, jr) 

12. /(x, y, z) = ye 3x + ze 2 -\ a = (0, 0, 2) 

13. /(jc, y, z) = xy- 3y 2 + 2xz, a = (2, -1, 1) 

14. /(x, z) = l/(x 2 + y 2 + z 2 + 1), a = (0, 0, 0) 

15. f(x, y, z) = sinxyz, a = (0, 0, 0) 

In Exercises 16-20, calculate the Hessian matrix ///(a) for 
the indicated function f at the indicated point a. 

16. f(x,y)= l/(x 2 + y 2 +l),a = (0,0) 

17. /(x, y) = cosx siny, a = (tt/4, it/3) 

18. f(x,y,z)= -£=,a = (l,2,-4) 

19. /(x, y, z) = x 3 + x 2 y - yz 2 + 2z\ a = (1, 0, 1) 

20. f(x, y, z) = e 2x ~^ sin5z, a = (0, 0, 0) 

21 . For / and a as given in Exercise 8, express the second- 
order Taylor polynomial pi(x, y), using the derivative 
matrix and the Hessian matrix as in formula (10) of 
this section. 

22. For / and a as given in Exercise 1 1 , express the second- 
order Taylor polynomial pi{x, y), using the derivative 
matrix and the Hessian matrix as in formula (10) of 
this section. 

23. For / and a as given in Exercise 12, express the second- 
order Taylor polynomial p2(x, y, z), using the deriva- 
tive matrix and the Hessian matrix as in formula (10) 
of this section. 



24. For / and a as given in Exercise 1 9, express the second- 
order Taylor polynomial pi(x, y, z), using the deriva- 
tive matrix and the Hessian matrix as in formula (10) 
of this section. 

25. Consider the function 

f(x u x 2 ,...,x,,) = e x > +2x > + - + " x ». 

(a) Calculate D/(0, 0, . . . , 0) and Hf(0, 0, . . . , 0). 

(b) Determine the first- and second-order Taylor poly- 
nomials of / at 0. 

(c) Use formulas (3) and ( 10) to write the Taylor poly- 
nomials in terms of the derivative and Hessian 
matrices. 

26. Find the third-order Taylor polynomial pi(x, y, z) of 

f(x,y,z) = e x+2y+3z 

at (0, 0, 0). 

27. Find the third-order Taylor polynomial of 

f{x, y, z) = x 4 + x 3 y + 2y 3 - xz 2 + x 2 y + 3xy - z + 2 

(a) at (0, 0, 0). 

(b) at (1,-1,0). 

Determine the total differential of the functions given in 
Exercises 28—32. 

28. /(x,y) = x 2 y 3 

29. f{x, y, z) = x 2 + 3y 2 - 2z 3 

30. fix, y, z) = cos (xyz) 

31. fix, y, z) = e x cosy + e y sin z 

32. fix, y, z) = l/^Jxyz 

33. Use the fact that the total differential df approximates 
the incremental change A/ to provide estimates of the 
following quantities: 

(a) (7.07) 2 (1.98) 3 

(b) l/y(4.1)(1.96)(2.05) 

(c) (1. 1) cos ((jr -0.03X0.12)) 

34. Near the point (1, —2, 1), is the function g(x, y, z) = 
x 3 — 2xy + x 2 z + 7z most sensitive to changes in x, 
y, or z? 

35. To which entry in the matrix is the value of the 
determinant 

2 3 
-1 5 

most sensitive? 



4.2 | Extrema of Functions 263 



36. If you measure the radius of a cylinder to be 2 in, with 
a possible error of ±0. 1 in, and the height to be 3 in, 
with a possible error of ±0.05 in, use differentials to 
determine the approximate error in 

(a) the calculated volume of the cylinder. 

(b) the calculated surface area. 

37. A can of mushrooms is currently manufactured to have 
a diameter of 5 cm and a height of 12 cm. The man- 
ufacturer plans to reduce the diameter by 0.5 cm. Use 
differentials to estimate how much the height of the 
can would need to be increased in order to keep the 
volume of the can the same. 

38. Consider a triangle with sides of lengths a and b that 
make an interior angle 0 . 

(a) If a = 3, b = 4, and 0 = tt/3, to changes in which 
of these measurements is the area of the triangle 
most sensitive? 

(b) If the length measurements in part (a) are in error 
by as much as 5% and the angle measurement is 
in error by as much as 2%, estimate the resulting 
maximum percentage error in calculated area. 

39. To estimate the volume of a cone of radius approx- 
imately 2 m and height approximately 6 m, how ac- 
curately should the radius and height be measured so 
that the error in the calculated volume estimate does 



not exceed 0.2 m 3 ? Assume that the possible errors in 
measuring the radius and height are the same. 

40. Suppose that you measure the dimensions of a block 
of tofu to be (approximately) 3 in by 4 in by 2 in. 
Assuming that the possible errors in each of your mea- 
surements are the same, about how accurate must your 
measurements be so that the error in the calculated 
volume of the tofu is not more than 0.2 in 3 ? What per- 
centage error in volume does this represent? 

41 . (a) Calculate the second-order Taylor polynomial for 

f(x, y) = cosx siny at the point (0, ?r/2). 

(b) If h = (h u h 2 ) = (x, y) - (0, tt/2) is such that 
\hi \ and \h 2 \ are no more than 0.3, estimate how 
accurate your Taylor approximation is. 

42. (a) Determine the second-order Taylor polynomial of 

f(x, y) = e x+2y at the origin. 

(b) Estimate the accuracy of the approximation if \x\ 
and \ y\ are no more than 0.1. 

43. (a) Determine the second-order Taylor polynomial of 

f(x, y) = e 2x cos y at the point (0, rr/2). 

(b) If h = (h u h 2 ) = (x, y) - (0, tt/2) is such that 
\h\\ < 0.2 and \h 2 \ < 0.1, estimate the accuracy 
of the approximation to / given by your Taylor 
polynomial in part (a). 



4.2 Extrema of Functions 

The power of calculus resides at least in part in its role in helping to solve a wide 
variety of optimization problems. With any quantity that changes, it is natural to 
ask when, if ever, does that quantity reach its largest, its smallest, its fastest or 
slowest? You have already learned how to find maxima and minima of a function 
of a single variable, and no doubt you have applied your techniques to a number of 
situations. However, many phenomena are not appropriately modeled by functions 
of only one variable. Thus, there is a genuine need to adapt and extend optimization 
methods to the case of functions of more than one variable. We develop the 
necessary theory in this section and the next and explore a few applications in §4.4. 

Critical Points of Functions 

Let X be open in R" and /:XcR"^Ra scalar- valued function. 



DEFINITION 2.1 We say that / has a local minimum at the point a in 
X if there is some neighborhood U of a such that f(x) > /(a) for all x 
in U. Similarly, we say that / has a local maximum at a if there is some 
neighborhood U of a such that /(x) < /(a) for all x in U. 



Figure 4.1 2 The graph of When n = 2, local extrema of f(x, y) are precisely the pits and peaks of the 

z = f(x, y). surface given by the graph of z = f(x, y), as suggested by Figure 4.12. 



Max. 




264 Chapter 4 I Maxima and Minima in Several Variables 



We emphasize our use of the adjective "local." When a local maximum of 
a function / occurs at a point a, this means that the values of / at points near 
a can be no larger, not that all values of / are no larger. Indeed, / may have 
local maxima and no global (or absolute) maximum. Consider the graphs in 
Figure 4.13. (Of course, analogous comments apply to local and global minima.) 




Figure 4.1 3 Examples of local and global maxima. 



Recall that, if a differentiable function of one variable has a local extremum 
at a point, then the derivative vanishes there (i.e., the tangent line to the graph 
of the function is horizontal). Figures 4.12 and 4.13 suggest strongly that, if a 
function of two variables has a local maximum or minimum at a point in the 
domain, then the tangent plane at the corresponding point of the graph must be 
horizontal. Such is indeed the case, as the following general result (plus formula 
(4) of §2.3) implies. 



THEOREM 2.2 Let X be open in R" and let /:XcR"^R be differentiable. 
If / has a local extremum at a 6 X, then Df(a) = 0. 



PROOF Suppose, for argument's sake, that / has a local maximum at a. Then the 
one-variable function F defined by F(t) = /(a + 1 h) must have a local maximum 
at t = 0 for any h. (Geometrically, the function F is just the restriction of / to the 
line through a parallel to h as shown in Figure 4. 14.) From one-variable calculus, 
we must therefore have F'(0) = 0. By the chain rule 



F'(t) = -[/(a + th)] = D/(a + th)h = V/(a + th) • h. 
at 









1 \ 
1 k 


1 / \ 


1 1 


1 / 1 


1 




1 J 







Graph of / 
restricted to line 




Figure 4.1 4 The graph of / restricted to a line. 



4.2 | Extrema of Functions 265 



Hence, 



/-o 

X/<o 

/ \ 

/ \ 
1 \ 


f-o 

v/ 
/\ 
/ \ 

/f> 0 \ 


\ f>0/ 


\ 1 
\ / 
\ / 


\ / 
\ / 




\ / 









Figure 4.1 5 The function / is 
strictly positive on the shaded 
region, strictly negative on the 
unshaded region, and zero along 
the lines y = ±x. 



0 = F'(0) = D/(a)h = f x ^)h x + / a (a)A 2 + • • • + A„(a)fc„. 

Since this last result must hold for all h e R", we find that by setting h in turn 
equal to (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1), we have 

f Xl {*)= = -•• = /*„ (a) = 0. 

Therefore, Df(a) = 0, as desired. ■ 

A point a in the domain of / where Df(a) is either zero or undefined is called 
a critical point of /. Theorem 2.2 says that any extremum of / must occur at a 
critical point. However, it is by no means the case that every critical point must 
be the site of an extremum. 

EXAMPLE 1 If f(x, y) = x 2 - y 2 , then Df(x, y) = [ 2x -2y ] so that, 
clearly, (0, 0) is the only critical point. However, neither a maximum nor a mini- 
mum occurs at (0, 0). Indeed, inside every open disk centered at (0, 0), no matter 
how small, there are points for which f(x, y) > f(0, 0) = 0 and also points where 
f(x, y) < /(O, 0). (See Figure 4.15.) ♦ 

This type of critical point is called a saddle point. Its name derives from the 
fact that the graph of z = f(x , y) looks somewhat like a saddle. (See Figure 4.16.) 




Figure 4.1 6 A saddle point. 



EXAMPLE 2 Let f(x, y) = ^x 1 + y 2 . The domain off is all of R 2 . We com- 

2x 2y 



pute that Df(x, y) 



; note that Df is unde- 



3( x 2 + y 2)2/3 3( x 2 + _ y 2)2/3 _ 

fined at (0, 0) and nonzero at all other (x, y) e R 2 . Hence, (0, 0) is the only 
critical point. Since f(x, y) > 0 for all (x, y) and has value 0 only at (0, 0), we 
see that / has a unique (global) minimum at (0, 0). ♦ 



The Nature of a Critical Point: The Hessian Criterion 

We illustrate our current understanding regarding extrema with the following 
example: 



EXAMPLE 3 We find the extrema of 

f(x, y) = x 2 + xy + y 2 + 2x- 2y + 5. 



266 Chapter 4 | Maxima and Minima in Several Variables 



/(«) 




Figure 4.17 An 

upward-opening parabola. 




Figure 4.18 A 

downward-opening parabola. 



Since / is a polynomial, it is differentiable everywhere, and Theorem 2.2 implies 
that any extremum must occur where df/dx and df/dy vanish simultaneously. 
Thus, we solve 



Bf 
dx 
Bf 
dy 



= 2x + y + 2 = 0 
= x+2y-2 = 0 



and find that the only solution is x = —2, y = 2. Consequently, (—2, 2) is the 
only critical point of this function. 

To determine whether (—2, 2) is a maximum or minimum (or neither), we 
could try graphing the function and drawing what we hope would be an obvious 
conclusion. Of course, such a technique does not extend to functions of more than 
two variables, so a graphical method is of limited value at best. Instead we'll see 
how / changes as we move away from the critical point: 

A/ = /(-2 + A,2 + *)-/(-2,2) 

= [(-2 + hf + (-2 + h)(2 + k) + {2 + kf 

+ 2(-2 + /!)-2(2 + £) + 5]- 1 
= h 2 + hk + k 2 . 

If the quantity Af = h 2 + hk + k 2 is nonnegative for all small values of h and k, 
then (—2, 2) yields a local minimum. Similarly, if Af is always nonpositive, then 
(—2, 2) must yield a local maximum. Finally, if Af is positive for some values 
of h and k and negative for others, then (—2, 2) is a saddle point. To determine 
which possibility holds, we complete the square: 



Af 



h 2 + hk + k 2 



h 2 + hk + 



,k + . k 



(h + \kf + Ik 2 . 



Thus, Af > 0 for all values of h and k, so (—2, 2) necessarily yields a local 
minimum. ♦ 

Example 3 with its attendant algebra clearly demonstrates the need for a better 
way of determining when a critical point yields a local maximum or minimum (or 
neither). In the case of a twice differentiable function /:XcR^R, you already 
know a quick method namely, consideration of the sign of the second derivative. 
This method derives from looking at the second-order Taylor polynomial of / 
near the critical point a, namely, 

f"(a) , 

f(x) « p 2 (x) = f(a) + / (a)(x -a)-\ — (x - a) 

= /(a)H ^—(x ~ a) , 

since /' is zero at the critical point a of /. If f"{a) > 0, the graph of y = pi(x) 
is an upward-opening parabola, as in Figure 4.17, whereas if f"{a) < 0, then 
the graph of y = pi(x) looks like the one shown in Figure 4.18. If f"(a) = 0, 
then the graph of y = pi(x) is just a horizontal line, and we would need to use 
a higher-order Taylor polynomial to determine if / has an extremum at a. (You 
may recall that when f"(a) = 0, the second derivative test from single-variable 
calculus gives no information about the nature of the critical point a.) 
The concept is similar in the context of n variables. Suppose that 

/(*)= f(xi,x 2 , . . . ,x n ) 



4.2 | Extrema of Functions 267 



is of class C 2 and that a = (a\, a 2 , . . . , a„) is a critical point of /. Then the 
second-order Taylor approximation to / gives 

a/ = /(x) - m « P2 (x) - m 

= Z)/(a)(x - a) + i(x - a) r ///(a)(x - a) 

when x a. (See Theorem 1.5 and formula (10) in §4.1.) Since / is of class C 2 
and a is a critical point, all the partial derivatives vanish at a, so that we have 
D/(a) = 0 and hence, 



A/«i(x-a) r ///(a)(x-a). 



(1) 



The approximation in (1) suggests that we may be able to see whether the in- 
crement A / remains positive (respectively, remains negative) for x near a and 
hence, whether / has a local minimum (respectively, a local maximum) at a by 
seeing what happens to the right side. 

Note that the right side of (1), when expanded, is quadratic in the terms 
(xj — cii). More generally, a quadratic form in h\, h 2 , ■ . . , h„ is a function Q 
that can be written as 

n 

Q(h u h 2 , h n ) = ^ bijhihj, 

where the bij 's are constants. The quadratic form Q can also be written in terms 
of matrices as 



Q(h)=[h l h 2 





~ b u 


b\ 2 ■ 


■ bl„ 




~ h x ~ 


h„] 


b 2 \ 


b 22 ■ 


■ b 2n 




h 2 




_ K\ 


bnl ■ 


■ h 

u nn - 




_ K _ 



h r fih, 



(2) 



where B = (pij). Note that the function Q is unchanged if we replace all bjj with 
^{bij + bjj). Hence, we may always assume that the matrix B associated to Q is 
symmetric, that is, that bjj = bj, (or, equivalently, that B T = B). Ignoring the 
factor of 1 /2, we see that the right side of (1) is the quadratic form in h = x — a, 
corresponding to the matrix B = Hf(a). 

A quadratic form Q (respectively, its associated symmetric matrix B) is said 
to be positive definite if Q(h) > 0 for all h ^ 0 and negative definite if Q(h) < 0 
for all h ^ 0. Note that if Q is positive definite, then Q has a global minimum (of 
0) at h = 0. Similarly, if Q is negative definite, then Q has a global maximum at 
h = 0. 

The importance of quadratic forms to us is that we can judge whether / has 
a local extremum at a critical point a by seeing if the quadratic form in the right 
side of ( 1) has a maximum or minimum at x = a. The precise result, whose proof 
is given in the addendum to this section, is the following: 



THEOREM 2.3 Let U c R" be open and /: U R a function of class C 2 . 
Suppose that a £ U is a critical point of /. 

1. If the Hessian Hf(a) is positive definite, then / has a local minimum at a. 

2. If the Hessian Hf(a) is negative definite, then / has a local maximum at a. 

3. If det Hf(a) 0 but Hf(a) is neither positive nor negative definite, then / 
has a saddle point at a. 



268 Chapter 4 I Maxima and Minima in Several Variables 



In view of Theorem 2.3, the issue thus becomes to determine when the Hessian 
Hf(a) is positive or negative definite. Fortunately, linear algebra provides an 
effective means for making such a determination, which we state without proof. 
Given a symmetric matrix B (which, as we have seen, corresponds to a quadratic 
form Q), let B^, for k = 1, . . . , n, denote the upper leftmost k x k submatrix of 
B. Calculate the following sequence of determinants: 

bn b 22 



det 5, 



det B 7 



det B 3 = 



bn b i2 bn 
bi\ bi2 bn 
b-i\ b i2 b 3i 



det£„ = det 5. 



If this sequence consists entirely of positive numbers, then B and Q are positive 
definite. If this sequence is such that det B^ < 0 for k odd and det B^ > 0 for k 
even, then B and Q are negative definite. Finally, if det B ^ 0, but the sequence 
of determinants det B\ , det B 2 , . . . , det B„ is neither of the first two types, then B 
and Q are neither positive nor negative definite. Combining these remarks with 
Theorem 2.3, we can establish the following test for local extrema: 



Second derivative test for local extrema. Given a critical point a of a func- 
tion / of class C 2 , look at the Hessian matrix evaluated at a: 



tf/(a) 



/*,*,(») fx lX2 (*) ■■■ /*,*„(») 
f X2 xM) fx 2 x 2 (a) ••• /x 2 x„(a) 



From the Hessian, calculate the sequence of principal minors of Hf(a). 
This is the sequence of the determinants of the upper leftmost square sub- 
matrices of Hf(a). More explicitly, this is the sequence d\, d 2 , . . . , d n , 
where dt = det Ht, and Hk is the upper leftmost k x k submatrix of Hf(a). 
That is, 

d\ 
di 



/x,x,(a), 






/««(») 


/v!* 2 (a) 




fx 2 xM) 


fx 2 x 2 (&) 


1 


fxixM) 


f xl xM 


f xl xM 


fx 2 xM) 


fx 2 xM 


fx 2 xM 


/v 3 *,(a) 


/v 3 * 2 (a) 


inu(a) 



di= fx 2 xM) fx 2 x 2 (a) fx 2 x 3 (a) , ...,d„ = \Hf(a)\. 



The numerical test is as follows: 
Assume that d„ = det Hf(a) ^ 0. 

1. If dk > 0 for k = 1 , 2, . . . , n, then / has a local minimum at a. 

2. If dk < 0 for k odd and dk > 0 for k even, then / has a local maximum 
at a. 

3. If neither case 1 nor case 2 holds, then / has a saddle point at a. 

In the event that det ///(a) = 0, we say that the critical point a is degenerate 
and must use another method to determine whether or not it is the site of an 
extremum of /. 



4.2 | Extrema of Functions 269 



EXAMPLE 4 Consider the function 



fxx 


fxy 




2 1 


. fyx 


fyy _ 




1 2 


minors 


is d 


i = /«(-2, 2) 



/(*, y) = x 2 + xy + y 2 + 2x - 2y + 5 

in Example 3. We have already seen that (—2, 2) is the only critical point. The 
Hessian is 



Hf(x, y) = 



\Hf(—2, 2)| = 3 (> 0). Hence, / has a minimum at (—2, 2), as we saw before, 
but this method uses less algebra. ♦ 

EXAMPLE 5 (Second derivative test for functions of two variables) Let us 

generalize Example 4. Suppose that f(x, y) is a function of two variables of class 
C 2 and further suppose that / has a critical point at a = (a, b). The Hessian 
matrix of / evaluated at {a, b) is 



Hf(a,b) 



f xx (a,b) f xy (a,b) 
f xy (a,b) f yy (a,b) 



Note that we have used the fact that f xy = f yx (since / is of class C 2 ) in con- 
structing the Hessian. The sequence of principal minors thus consists of two 
numbers: 

d\ = f xx (a,b) and d 2 = f xx (a, b)f yy (a, b) - f xy (a, bf . 
Hence, in this case, the second derivative test tells us that 

1. / has a local minimum at (a, b) if 

f xx (a, b) > 0 and f xx (a, b)f yy (a, b) - f xy (a, bf > 0. 

2. / has a local maximum at {a, b) if 

f xx (a, b) < 0 and f xx (a, b)f yy (a, b) - f xy (a, bf > 0. 

3. / has a saddle point at (a, b) if 

fxxia, b)f yy (a, b) - f xy (a, bf < 0. 

Note that if f xx (a, b)f yy (a, b) — f xy (a, bf = 0, then / has a degenerate critical 
point at (a, b) and we cannot immediately determine if {a, b) is the site of a local 
extremum of /. ♦ 

EXAMPLE 6 Let f(x, y, z) = x 3 + xy 2 + x 2 + y 2 + 3z 2 . To find any local 
extrema of /, we must first identify the critical points. Thus, we solve 

Df(x,y,z)= [3x 2 + y 2 + 2x 2xy + 2y 6z] = [0 0 0] . 

From this, it is not hard to see that there are two critical points: (0, 0, 0) and 
(-§, 0,0). The Hessian of/ is 



Hf(x,y,z) = 



6x + 2 

2y 
o 



2y 
2x + 2 
0 



270 Chapter 4 I Maxima and Minima in Several Variables 



At the critical point (0, 0, 0), we have 



Hf(0, 0, 0) = 



2 0 0 
0 2 0 
0 0 6 



and its sequence of principal minors is d\ =2, d 2 = 4, dj = 24. Since these 
determinants are all positive, we conclude that / has a local minimum at (0,0,0). 
At (-§,0,0), we calculate that 



ff/(--,0,0 



-2 0 0 
0 | 0 
0 0 6 



The sequence of minors is —2, — |, — 8. Hence, / has a saddle point at (— |, 0, 0). 







+ 


h>0 






+ 








+ 








+ 




1 — 

-1 


0 


+ 


1 

1 






+ 








+ 








+ 





Figure 4.1 9 Away from the 
origin, the function h of Example 7 
is negative along the x-axis and 
positive along the _v-axis. 



EXAMPLE 7 To get a feeling for what happens in the case of a degenerate 
critical point (i.e., a critical point a such that det Hf(a) = 0), consider the three 
functions 



/(x,y) = x 4 + x 2 + y 4 , 



g(x, y) = -x 4 - x 2 



and 



h(x, y) = x 



x 2 + y 4 . 



We leave it to you to check that the origin (0, 0) is a degenerate critical point 
of each of these functions. (In fact, the Hessians themselves look very similar.) 
Since / is 0 at (0, 0) and strictly positive at all (x, y) ^ (0, 0), we see that / 
has a strict minimum at the origin. Similar reasoning shows that g has a strict 
maximum at the origin. For h, the situation is slightly more complicated. Along 
the y-axis, we have h(0, y) = y 4 , which is zero at y = 0 (the origin) and strictly 
positive everywhere else. Along the x-axis, 



h(x,0) = x 4 



x 2 = x 2 (x 



1)(JC + 1). 



For — 1 < x < 1 and x ^ 0, h(x, 0) < 0. We have the situation depicted in Fig- 
ure 4. 19. Thus, every neighborhood of (0, 0) contains some points (x, y) where h 
is positive and also some points where h is negative. Therefore, h has a saddle point 
at the origin. The "moral of the story" is that a degenerate critical point can exhibit 
any type of behavior, and more detailed consideration of the function itself, rather 
than its Hessian, is necessary to understand its nature as a site of an extremum. ♦ 



Global Extrema on Compact Regions 

Thus far our discussion has been limited to consideration of only local extrema. 
We have said nothing about how to identify global extrema, because there really is 
no general, effective method for looking at an arbitrary function and determining 
whether and where it reaches an absolute maximum or minimum value. For the 
purpose of applications, where finding an absolute maximum or minimum is 
essential, such a state of affairs is indeed unfortunate. Nonetheless, we can say 
something about global extrema for functions defined on a certain type of domain. 



4.2 | Extrema of Functions 271 




Figure 4.20 Compact regions. 



DEFINITION 2.4 A subset X C R" is said to be compact if it is both closed 
and bounded. 



Recall that X is closed if it contains all the points that make up its boundary. 
(See Definition 2.3 of §2.2.) To say that X is bounded means that there is some 
(open or closed) ball B that contains it. (That is, X is bounded if there is 
some positive number M such that ||x|| < M for all x e X.) Thus, compact sets 
contain their boundaries (a consequence of being closed) and have only finite 
extent (a consequence of being bounded). Some typical compact sets in R 2 and 
R 3 are shown in Figure 4.20. 

For our purposes the notion of compactness is of value because of the next 
result, which we state without proof. 



THEOREM 2.5 (Extreme value theorem) If X c R" is compact and /: 
X -> R is continuous, then / must have both a global maximum and a global 
minimum somewhere on X. That is, there must exist points a max and a m ; n in X 
such that, for all x e X, 

/(a m in) < /(x) < /(a max ). 



We need the compactness hypothesis since a function defined over a noncom- 
pact domain may increase or decrease without bound and hence, fail to have any 
global extremum, as suggested by Figure 4.21. This is analogous to the situation 
in one variable where a continuous function defined on an open interval may fail 
to have any extrema, but one defined on a closed interval (which is a compact 
subset of R) must attain both maximum and minimum values. (See Figure 4.22.) 
In the one-variable case, extrema can occur either in the interior of the interval 
or else at the endpoints. Therefore, you must compare the values of / at any 
interior critical points with those at the endpoints to determine which is largest 
and smallest. In the case of functions of n variables, we do something similar, 
namely, compare the values of / at any critical points with values at any restricted 
critical points that may occur along the boundary of the domain. 



272 Chapter 4 I Maxima and Minima in Several Variables 




x 



Figure 4.21 A graph that lacks a 

global minimum. Figure 4.22 The function depicted by the graph on the left has no global extrema — the 

function is defined on the open interval (a, b). By contrast, the function defined on the 
closed interval [a, b], and with the graph on the right, has both a global maximum and 
minimum. 



EXAMPLE 8 LetTiXcR 2 





y = 2 




x = -l 






x = 2 












y = -l 





R be given by 



Figure 4.23 The domain of the 
function T of Example 8. 



T(x, y) = x 2 -xy + y 2 + 1, 

where X is the closed square in Figure 4.23. (Note that X is compact.) Think of 
the square as representing a fiat metal plate and the function T as the temperature 
of the plate at each point. Finding the global extrema amounts to finding the 
warmest and coldest points on the plate. According to Theorem 2.5, such points 
must exist. 

We need to find all possible critical points of T. Momentarily considering T 
as a function on all of R 2 , we find the usual critical points by setting DT(x, y) 
equal to 0. The result is the system of two equations 

| 2x - y = 0 
|-x + 2y = 0' 

which has (0, 0) as its only solution. Whether it is a local maximum or minimum 
is not important for now, because we seek global extrema. Because there is only 
one critical point, at least one global extremum must occur along the boundary of 
X (which consists of the four edges of the square). We now find all critical points 
of the restriction of T to this boundary: 

1. The bottom edge of X is the set 

E l ={(x,y)\y = -l,-l<x <2}. 

The restriction of T to E\ defines a new function f\ : [— 1 , 2] — > R given by 

f x (x)= T(x,-\) = x 2 +x + 2. 

As f{(x) = 2x + 1, the function f\ has a critical point at x = — j. Thus, 
we must examine the following points of X for possible extrema: (— \ , — l), 
(— 1 , — 1), and (2, — 1). (The first point is the critical point of f\ , and the second 
two are the vertices of X that lie on E\ .) 

2. The top edge of X is given by 



E 2 = {(x,y) \ y = 2, -1 <x < 2}. 



4.2 | Extrema of Functions 273 



Consequently, we define fy. [— 1 , 2] -» R by 

/ 2 (jc) = T(jr, 2) = x 2 - 2x + 5. 

(/2 is the restriction of 7 1 to £2-) We calculate f^ipc) = 2x — 2, which implies 
that x = 1 is a critical point of /2. Hence, we must consider (1, 2), (—1, 2), 
and (2, 2) as possible sites for global extrema of T. (The points (—1,2) and 
(2, 2) are the remaining two vertices of X.) 

3. The left edge of X is 

E 3 = {(x,y)\x = -l,-\ < v<2}. 

Therefore, we define fa : [— 1 , 2] -> R by 

/ 3 (y)=r(-l,y) = y 2 + y + 2. 

We have / 3 '(y) = 2y + 1, and so y = — I is the only critical point of fa. Thus 
(— 1 , — 2) is a potential site of a global extremum. (We need not worry again 
about the vertices (—1,-1) and (— 1 , 2).) 

4. The right edge of X is 

£ 4 = {(x,y)|x=2,-l <y<2}. 

We define / 4 : [-1,2] -> Rby 

f 4 (y) = 7X2, y) = y 2 - 2y + 5. 

We have / 4 '(y) = 2y — 2, and so y = 1 is the only critical point of / 4 . Hence, 
we must include (2, 1) in our consideration. 

Consequently, we have nine possible locations for global extrema, shown in 
Figure 4.24. Now we need only to compare the actual values of T at these points 
to see that (0, 0) is the coldest point on the plate and both (2, — 1) and (—1,2) are 
the hottest points. ♦ 

If a function is defined over a noncompact region, there is no general result 
like the extreme value theorem (Theorem 2.5) to guarantee existence of any global 
extrema. However, ad hoc arguments frequently can be used to identify global 
extrema. 



(-1,2) 



(1.2) (2,2) 



(0,0) 



(-1,-1)' 
(-1,-1)/ 

H,-i) 



(2,1) 



(2,-1) 



(*o0 


T(x,y) 


(0,0) 


1 


(-5,-1) 


7 
4 


(-1,-1) 


2 


(2,-1) 


8 


(1,2) 


4 


(-1,2) 


8 


(2,2) 


5 


(-1,-1) 


7 
4 


(2,1) 


4 



Figure 4.24 Possible global extrema for T . 



274 Chapter 4 I Maxima and Minima in Several Variables 



EXAMPLE 9 Consider the function f(x, y) = e 1 ' 3 ^ denned on all of R 2 
(so the domain is certainly not compact). Verifying that / has a unique critical 
point at (0, 0) is straightforward. We leave it to you to check that the Hessian 
criterion implies that / has a local maximum there. In any case, for all (x,y) 6 R 2 , 
we have 

1 - 3x 2 - y 2 < 1. 

Therefore, because the exponential function is always increasing (i.e., if u\ < u%, 
thene" 1 < e" 2 ), 

As f(0, 0) = e, we see that / has a global maximum at (0, 0). ♦ 

WARNING It is tempting to assume that if a function has a unique critical point 
that is a local extremum, then it must be a global extremum as well. Although 
true for the case of a function of a single variable, it is not true for functions of 
two or more variables. (See Exercise 52 for an example.) 

Addendum: Proof of Theorem 2.3 

Step 1. We show the following key property of a quadratic form Q, namely, that 
if k e R, then 

Q(Xh) = X 2 Q(h). (3) 

This is straightforward to establish if we write Q in terms of its associated sym- 
metric matrix B and use some of the properties of matrix arithmetic given in § 1 .6: 

g(A.h) = (Xh) T B(Xh) = kh T B(Xh) = X 2 h T Bh = A 2 g(h). 

Step 2. We show that if B is the symmetric matrix associated to a positive 
definite quadratic form Q, then there is a positive constant M such that 

e(h) > Mpii 2 

for allh 6 R". 

First, note that when h = 0, then Q(h) = <2(0) = 0 so the conclusion holds 
trivially in this case. 

Next, suppose that his a unit vector (i.e., ||h|| = 1). The (endpoints of the) set 
of all unit vectors in R" is an (n — l)-dimensional sphere S, which is a compact 
set. Hence, by the extreme value theorem (Theorem 2.5), the restriction of Q to 
S must achieve a global minimum value M somewhere on S. Thus, Q(h) > M 
for all h € S. 

Finally, let h be any nonzero vector in R". Then its normalization h/||h|| is a 
unit vector and so lies in S. Therefore, by the result of Step 1, we have 

G(h)=e(l|l.|l 1 | | |) = l|hfG( 1 | I )>IIN 2 «, 

since h/||h|| is in S. 

Step 3. Now we prove the theorem. By the second-order Taylor formula 
Theorem 1.5 and formula (10) of §4.1, we have that, for the critical point a of /, 

Af = f(x) - /(a) = i(x - a) r ///(a)(x - a) + R 2 (x, a), (4) 
where |i?2(x, a)|/||x — a|| 2 -> 0 as x -» a. 



4.2 | Extrema of Functions 275 



Suppose first that Hf(&) is positive definite. Then by Step 2 with h = x — a, 
there must exist a constant M > 0 such that 

i(x-a) r ///(a)(x-a)> M||x-a|| 2 . (5) 

Because |i? 2 (x, a)|/||x — a|| 2 -> 0 as x -> a, there must be some 5 > 0 so that if 
0 < ||x - a|| < <5, then |i?2(x, a)|/||x - a|| 2 < M, or, equivalently, 

|i? 2 (x,a)| <M||x-a|| 2 . (6) 

Therefore, (4), (5), and (6) imply that, for 0 < ||x — a|| < <5, 

A/>0 

so that / has a (strict) local minimum at a. 

If Hf(a) is negative definite, then consider g = —f. We see that a is also a 
critical point of g and that Hg(a) = — Hf(a), so Hg(a) is positive definite. Hence, 
the argument in the preceding paragraph shows that g has a local minimum at a, 
so / has a local maximum at a. 

Now suppose det Hf(a) 0, but that Hf(a) is neither positive nor negative 
definite. Let xi be such that 

i(x, - a) r ///(a)( Xl - a) > 0 

and X2 such that 

i(x 2 - a) r ///(a)(x 2 - a) < 0. 

(Since det Hf(a) 0, such points must exist.) For i = 1,2 let 

y t (0 = t(x - a) + a, 

the vector parametric equation for the line through a and x,- . Applying formula 
(4) with x = y,-(f), we see 

A/ = f(Yi(t)) - f(a) = i(y,(f) - a) r H/(a)(y,(0 " a) + ^(y,(0, a) 
= i(y ( - 0 - a) r Jf/(aXy,(0 - a) + ||y,(f) - a|| 2 g / 

lly/(0-a|| 2 

Note that y, (f) — a = f (x, — a). Therefore, using the property of quadratic forms 
given in Step 1 and the fact that ||y,(?) - a|| 2 = ||?(x ; - a)|| 2 = r 2 ||x/ - a|| 2 , we 
have 

/(y (0) - /(a) 

(7) 



= t 2 



i(x ; - tfHfW* - a) + ||x, - a|| 2 f 2 ^ (0 ' ^ 

llyi(0 - all 2 



Now note that, for i = 1 , the first term in the brackets in the right side of (7) is a 
positive number P and, for i = 2, it is a negative number N. Set 



M = min 



|X! - a|| 2 ' ||x 2 -a|| 2 



Because we know that \R z (yi(t), a)|/||y(r) - a|| 2 -> 0 as t -+ 0, we can find 
some S > 0 so that if 0 < t < S, then 

1*2(7,(0, a)| <M 



|y(0 -a|| 5 



276 Chapter 4 [ Maxima and Minima in Several Variables 



But this implies that, for 0 < t < S, 

A/ = /(yi(O)-/(a)>0, 

while 

A/ = /(y 2 (0) " /(*) < 0. 
Thus, / has a saddle point at x = a. 



4.2 Exercises 



1. Concerning the function f(x, y) = Ax + 6y — 12 — 

2 2 

x — y : 

(a) There is a unique critical point. Find it. 

(b) By considering the increment Af, determine 
whether this critical point is a maximum, a mini- 
mum, or a saddle point. 

(c) Now use the Hessian criterion to determine the 
nature of the critical point. 

2. This problem concerns the function g(x, y) = x 2 — 
2y 2 + 2x + 3. 

(a) Find any critical points of g. 

(b) Use the increment Ag to determine the nature of 
the critical points of g. 

(c) Use the Hessian criterion to determine the nature 
of the critical points. 

In Exercises 3-20, identify and determine the nature of the 
critical points of the given functions. 



3. 


/(*. y) = 


2xy - 2x 2 - 5y 2 + Ay - 3 


4. 


fix, y) = 


ln(x 2 + y 2 + 1) 


5. 


fix, y) = 


x 2 + y 3 — 6xy + 3x + 6y 


6. 


fix, y) = 


y 4 — 2xy 2 + x 3 — x 


7. 


fix, y) = 


8 1 

XV + - + - 

x y 


8. 


fix, y) = 


e x sin y 


9. 


fix, y) = 


e -y(x 2 - y 2 ) 


10. 


fix, y) = 


{X + y)(l - xy) 


11. 


fix, y) = 


x — y — x y + y 


12. 


fix,y) = 


e~ x (x 2 + 3y 2 ) 


13. 


fix, y) = 


2x — 3y + Inxy 


14. 


fix, y) = 


cosx siny 


15. 


fix,y,z) 


= x 2 — xy + z 2 — 2xz + 6z 


16. 


fix, y, z) 


= (x 2 +2y 2 + l)cosz 


17. 


fix, y, z) 


= x 2 + y 2 + 2z 2 + xz 



18. f(x, y, z) = jc 3 + xz 2 - 3x 2 + y 2 + 2z 2 

1 

19. f(x,y,z) = xy + xz + 2yz+ - 

x 

20. f(x,y,z) = e x (x 2 - y 2 - 2z 2 ) 

21. (a) Find all critical points of fix,y) = 

2y 3 - 3y 2 - 36y + 2 
1 + 3x 2 ' 
(b) Identify any and all extrema of /. 

22. (a) Under what conditions on the constant k will the 

function 

f{x, y) = kx 2 — 2xy + ky 2 

have a nondegenerate local minimum at (0, 0)? 
What about a local maximum? 

(b) Under what conditions on the constant k will the 
function 

g(x, y, z) = kx 2 + kxz — 2yz — y 2 + -Z 2 

have a nondegenerate local maximum at (0, 0, 0)? 
What about a nondegenerate local minimum? 

23. (a) Consider the function f(x, y) = ax 2 + by 2 , 

where a and b are nonzero constants. Show that 
the origin is the only critical point of /, and deter- 
mine the nature of that critical point in terms of a 
and b. 

(b) Now consider the function f(x,y,z) = ax 2 + 
by 2 + cz 2 , where a, b, and c are all nonzero. Show 
that the origin in R 3 is the only critical point of /, 
and determine the nature of that critical point in 
terms of a, b, and c. 

(c) Finally, let fix\, x%, x„) = a\x\ + 02X 2 
+ ■ ■ ■ + a,,x 2 , where a; is a nonzero constant for 
i = 1, 2, . . . , n. Show that the origin in R" is the 
only critical point of /, and determine its nature. 

Sometimes it can be difficult to determine the critical point of 
a function f because the system of equations that arises from 
setting V/ equal to zero may be very complicated to solve by 
hand. For the functions given in Exercises 24—27, (a) use a 
computer to assist you in identifying all the critical points of 
the given function f , and (b) use a computer to construct the 



4.2 I Exercises 277 



Hessian matrix and determine the nature of the critical points 
found in part (a). 

^ 24. f(x, y) = y 4 + x 3 - 2xy 2 - x 
^> 25. f(x, y) = 2x 3 y - y 2 - 3xy 

^ 26. f(x, y, z ) = yz- xyz - x 2 - y 2 - 2z 2 

27. f(x, y, z, u>) = yw — xyz — x 2 — 2z 2 + w 2 

28. Show that the largest rectangular box having a fixed 
surface area must be a cube. 

29. What point on the plane 3x — 4y — z = 24 is closest 
to the origin? 

30. Find the points on the surface xy + z 2 = 4 that are 
closest to the origin. Be sure to give a convincing ar- 
gument that your answer is correct. 

31 . Suppose that you are in charge of manufacturing two 
types of television sets. The revenue function, in dol- 
lars, is given by 

R(x, y) = 8x + 6y - x 2 - 2y 2 + 2xy, 

where x denotes the quantity of model X sets sold, and 
y the quantity of model Y sets sold, both in units of 
100. Determine the quantity of each type of set that 
you should produce in order to maximize the resulting 
revenue. 

32. Find the absolute extrema of f(x, y) = x 2 + xy + 
y 2 — 6y on the rectangle {(x, y) \ — 3 < x < 3, 0 < 

y < 5}'. 

33. Find the absolute maximum and minimum of 

fix, y, z) = x 2 + xz — y 2 + 2z 2 + xy + 5x 

on the block {(x, y,z)\ - 5 < x < 0, 0 < y < 3, 
0 < z < 2}. 

34. A metal plate has the shape of the region x 2 + y 2 < 1 . 
The plate is heated so that the temperature at any point 
(x , y) on it is indicated by 

T(x,y) = 2x 2 + y 2 -y + 3. 

Find the hottest and coldest points on the plate and the 
temperature at each of these points. (Hint: Parametrize 
the boundary of the plate in order to find any critical 
points there.) 

35. Find the (absolute) maximum and minimum values of 
f(x, y) = sinx cosy on the square R = {(x, y) | 0 < 
x < 2it, 0 < y < 2jt}. 

36. Find the absolute extrema of f(x, y) = 2 cosx + 
3 siny on the rectangle {(x, y) \ 0 < x < 4, 0 < 

y <3}. 

37. Determine the absolute minimum and maximum 
values of the function f(x, y) = 2x 2 — 2xy + y 2 



— y + 3 on the closed triangular region with vertices 
(0, 0), (2, 0), and (0, 2). 

38. Determine the absolute minimum and maximum val- 
ues of the function f(x, y) = x 2 y on the elliptical re- 
gion D = {(x, y) | 3x 2 + 4y 2 < 12}. 

39. Find the absolute extrema of f(x, y, z) = 
e !- x 2_ y 2 +2y _ z 2_ 4z Qn ^ ball {( ^ _^ z ) | x 2 + y 2 - 2y 

+ z 2 + 4z < 0}. 

Each of the functions in Exercises 40-45 has a critical point 
at the origin. For each function, (a) check that the Hessian 
fails to provide any information about the nature of the critical 
point at the origin, and (b) find another way to determine if the 
function has a maximum, minimum, or neither at the origin. 

40. f(x,y) = x 2 y 2 

41. f(x,y) = 4-3x 2 y 2 

42. f(x,y) = x 3 y 3 

43. f(x, y, z) = x 2 y 3 z 4 

44. f(x, y, z) = x 2 y 2 z 4 

45. f(x,y,z) = 2-x 4 y 4 -z 4 

In Exercises 46—48, (a) find all critical points of the given 
function f and identify their nature as local extrema and (b) 
determine, with explanation, any global extrema of f. 

46. f(x,y) = e x2+5 > 2 

47. f(x,y, z) = e 2 -* 2 - 2 } 2 - 3 '- 4 

48. f(x, y) = x 3 + y 3 - 3xy + 7 

49. Determine the global extrema, if any, of 

f(x, y) = xy + 2y — lnx — 2 lny, 
where x, y > 0. 

50. Find all local and global extrema of the function 

f{x, y, z) = x 3 + 3x 2 + e> ,2+l +z 2 - 3xz. 

51. Let f(x, y) = 3 - [(x - l)(y - 2)] 2 / 3 . 

(a) Determine all critical points of /. 

(b) Identify all extrema of /. 

52. (a) Suppose /: R — > R is a differentiate function of 

a single variable. Show that if / has a unique crit- 
ical point at xo that is the site of a strict local ex- 
tremum of /, then / must attain a global extremum 
at xo. 

(b) Let fix, y) = 3ye x - e 3x - y 3 . Verify that / has 
a unique critical point and that / attains a local 
maximum there. However, show that / does not 
have a global maximum by considering how / be- 
haves along the y-axis. Hence, the result of part 
(a) does not carry over to functions of more than 
one variable. 



278 Chapter 4 I Maxima and Minima in Several Variables 



53. (a) Let / be a continuous function of one variable. 
Show that if / has two local maxima, then / must 
also have a local minimum. 

(b) The analogue of part (a) does not necessarily hold 
for continuous functions of more than one variable, 



as we now see. Consider the function 



f(x,y) = 2-(xy 2 



D 2 -(v 2 -l) 2 



Show that / has just two critical points — and that 
both of them are local maxima. 

(c) Use a computer to graph the function / in part (b). 



4.3 Lagrange Multipliers 

Constrained Extrema 

Frequently, when working with applications of calculus, you will find that you 
do not need simply to maximize or minimize a function but that you must do so 
subject to one or more additional constraints that depend on the specifics of the 
situation. The following example is a typical situation: 

EXAMPLE 1 An open rectangular box is to be manufactured having a (fixed) 
volume of 4 ft 3 . What dimensions should the box have so as to minimize the 
amount of material used to make it? 

We'll let the three dimensions of the box be independent variables x, y, and 
z, shown in Figure 4.25. To determine how to use as little material as possible, 
we need to minimize the surface area function A given by 

A(x, y, z) = 2xy + 2yz + xz 

front and back sides bottom only 

For x, y, z > 0, this function has neither minimum nor maximum. However, we 
have not yet made use of the fact that the volume is to be maintained at a constant 
4 ft 3 . This fact provides a constraint equation, 

V(x, y, z) = xyz = 4. 

The constraint is absolutely essential if we are to solve the problem. In particular, 
the constraint enables us to solve for z in terms of x and y: 

4 

xy' 

We can thus create a new area function of only two variables: 

/ 4 

a(x, y) = A j x, y, — 
V xy 

8 4 
= 2xy+ - + -. 

x y 

Now we can find the critical points of a by setting Da equal to 0: 



v 



Figure 4.25 The open box of 
Example 1. 



4.3 | Lagrange Multipliers 



The first equation implies 



so that the second equation becomes 



2x - 4 



16 



or, equivalently, 



x 1 - -x* = 0 



The solutions to this equation are x = 0 (which we reject) and x = 2. Thus, the 
critical point of a of interest is (2, 1), and the constrained critical point of the 
original function A is (2, 1,2). 

We can use the Hessian criterion to check that x = 2, y = 1 yields a local 
minimum of a : 



Ha(x, y) = 



16/x 3 
2 



2 

8/y 3 



so Ha(2, 1) = 



The sequence of minors is 2, 12 so we conclude that (2, 1) does yield a local 
minimum of a. Because a(x, y) — >■ oo as either jc — > 0 + , y — > 0 + , je — > oo, or 
y -> oo, we conclude that the critical point must yield a global minimum as well. 
Thus, the solution to the original question is to make the box with a square base 
of side 2 ft and a height of 1 ft. ♦ 



The abstract setting for the situation discussed in Example 1 is to find max- 
ima or minima of a function f(x\, xi, ■ ■ ■ , x n ) subject to the constraint that 
g{x\, X2, ■ ■ ■ , x n ) = c for some function g and constant c. (In Example 1, the 
function / is A(x, y, z), and the constraint is xyz = 4.) One method for finding 
constrained critical points is used implicitly in Example 1: Use the constraint 
equation g(x) = c to solve for one of the variables in terms of the others. Then 
substitute for this variable in the expression for /(x), thereby creating a new 
function of one fewer variables. This new function can then be maximized or 
minimized using the techniques of §4.2. In theory, this is an entirely appropriate 
way to approach such problems, but in practice there is one major drawback: It 
may be impossible to solve explicitly for any one of the variables in terms of the 
others. For example, you might wish to maximize 

f(x,y,z) = x 2 + 3y 2 + y 2 z 4 



subject to 



g(x,y,z) = e x 



x 5 y 2 z + cos 



= 2. 



There is no means of isolating any of x, y, or z on one side of the constraint 
equation, and so it is impossible for us to proceed any further along the lines of 
Example 1. 



280 Chapter 4 | Maxima and Minima in Several Variables 



The Lagrange Multiplier 

The previous discussion points to the desirability of having another method for 
solving constrained optimization problems. The key to such an alternative method 
is the following theorem: 



THEOREM 3.1 Let X be open in R" and f,g:X^Rbe functions of class C 1 . 
Let S = {x e X \ g(x) = c} denote the level set of g at height c. Then if / \ s (the 
restriction of / to S) has an externum at a point x 0 eS such that Vg(x 0 ) 0, 
there must be some scalar X such that 

V/(x 0 ) = AVg(x 0 ). 



The conclusion of Theorem 3.1 implies that to find possible sites for extrema 
of / subject to the constraint that g(x) = c, we can proceed in the following 
manner: 

1. Form the vector equation V/(x) = XVg(x). 

2. Solve the system 

V/(x) = AVg(x) 
g(X) = c 

for x and X. When expanded this is actually a system of n + 1 equations in 
n + 1 unknowns x\, x 2 , ■ ■ ■ , x n , X, namely, 

f Xl (xi,x 2 , ■ ■ .,x„) = kg Xl (xi,x 2 , ■ ■ -,x„) 
f X2 (xi,x 2 , ...,x n ) = Xg X2 (xi,x 2 , ■ ■ ■ , x n ) 

• 

fx„( x i >x 2 ,...,x n ) = Xg Xi Xxi ,x 2 , x n ) 
g(x u x 2 , . . . ,x„) = c 

The solutions for x = (x\, x 2 , . . . , x n ) in the system above, along with any 
other points x satisfying the constraint g(x) = c and such that V/ is unde- 
fined or Vg vanishes or is undefined, are the candidates for extrema for the 
problem. 

3. Determine the nature of / (as maximum, minimum, or neither) at the critical 
points found in Step 2. 

The scalar A appearing in Theorem 3. 1 is called a Lagrange multiplier, after 
the Italian-born French mathematician Joseph-Louis Lagrange (1736-1813) who 
first developed this method for solving constrained optimization problems. In 
practice, Step 2 can involve some algebra, so it is important to keep your work 
organized. (Alternatively, you can use a computer to solve the system.) In fact, 
since the Lagrange multiplier X is usually not of primary interest, you can avoid 
solving for it explicitly, thereby reducing the algebra and arithmetic somewhat. 
Determining the nature of a constrained critical point (Step 3) can be a tricky 
business. We'll have more to say about that issue in the examples and discussions 
that follow. 

EXAMPLE 2 Let us use the method of Lagrange multipliers to identify the 
critical point found in Example 1 . Thus, we wish to find the minimum of 

A(x, y, z) = 2xy + 2yz + xz 



4.3 | Lagrange Multipliers 281 



subject to the constraint 

V(x, y, z) = xyz = 4. 
Theorem 3.1 suggests that we form the equation 

VA(x,y,z) = XVV(x,y,z). 

This relation of gradients coupled with the constraint equation gives rise to the 
system 

2y + z = kyz 
2x + 2z = ~kxz 
2y + x = kxy 
xyz = 4 

Since A is not essential for our final solution, we can eliminate it by means of any 
of the first three equations. Hence, 

_ 2y + z _2x +2z _2y + x 

yz xz xy 

Simplifying, this implies that 

2 1 2 2 2 1 
- + - = - + - = - + -. 

z y z x x y 

The first equality yields 

— = — or x = 2y, 

y x 

while the second equality implies that 

2 1 

- = - or z = 2y. 

z y 

Substituting these relations into the constraint equation xyz = 4 yields 

(2y)(y)(2y) = 4, 

so that we find that the only solution isy = l,je=z = 2, which agrees with our 
work in Example 1. (Note that V V = 0 only along the coordinate axes, and such 
points do not satisfy the constraint V(x, y, z) = 4.) ♦ 



An interesting consequence of Theorem 3. 1 is this: By Theorem 6.4 of Chap- 
ter 2, we know that the gradient Vg, when nonzero, is perpendicular to the level 
sets of g. Thus, the equation V/ = XV g gives the condition for the normal vector 
to a level set of / to be parallel to that of a level set of g. Hence, for a point xo 
to be the site of an extremum of /' on the level set S = {x | g(x) = c], where 
Vg(x 0 ) ^ 0, we must have that the level set R of / that contains x 0 is tangent to 
S at xo. 

EXAMPLE 3 Consider the problem of finding the extrema of f(x, y) = x 2 /4 
+ y 2 subject to the condition that x 2 + y 2 = 1. We let g(x, y) = x 2 + y 2 , and 
so the Lagrange multiplier equation V f(x, y) = XVg(x, y), along with the 



282 Chapter 4 | Maxima and Minima in Several Variables 



constraint equation, yields the system 

x 




Figure 4.26 The level sets of the 
function f(x, y) = x 2 /4 + y 2 
define a family of ellipses. The 
extrema of / subject to the 
constraint that x 2 + y 2 = 1 (i.e., 
that lie on the unit circle) occur at 
points where an ellipse of the 
family is tangent to the unit circle. 



= 2Xx 



2y = 2Xy ■ 

x 2 + y 2 = 1 

(There are no points simultaneously satisfying g(x, y) = 1 and Vg(x,y) = 
(0, 0).) The first equation of this system implies that either x = 0 or X = \. 
If x = 0, then the second two equations, taken together, imply that y = ±1 and 
X = 1 . If X = \, then the second two equations imply y = 0 and x — ±1 . There- 
fore, there are four constrained critical points: (0, ±1), corresponding to X = 1, 
and (±1, 0), corresponding to X = |. 

We can understand the nature of these critical points by using geometry and 
the preceding remarks. The collection of level sets of the function / is the family 
of ellipses x 2 /4 + y 2 = k whose major and minor axes lie along the x- and y- 
axes, respectively. In fact, the value f(x, y) = x 2 /A + y 2 = k is the square of 
the length of the semiminor axis of the ellipse x 2 /4 + y 2 = k. The optimization 
problem then is to find those points on the unit circle x 2 + y 2 = 1 that, when 
considered as points in the family of ellipses, minimize and maximize the length 
of the minor axis. When we view the problem in this way, we see that such points 
must occur where the circle is tangent to one of the ellipses in the family. A sketch 
shows that constrained minima of / occur at (± 1 , 0) and constrained maxima at 
(0, ± 1 ). In this case, the Lagrange multiplier X represents the square of the length 
of the semiminor axis. (See Figure 4.26.) ♦ 

EXAMPLE 4 Consider the problem of determining the extrema of f(x, y) = 
2x + y subject to the constraint that *Jx + Jy = 3. We let g(x, y) = *Jx + ^/y, 
so that the Lagrange multiplier equation V f(x, y) = XVg(x, y), along with the 
constraint equation, yields the system 

X 



2jx 
X 

1 = ■ 

yfx + s/y = 3 

The first two equations of this system imply that X = 4^/x = 2Jy so that Jy = 
2yfx. Using this in the last equation, we find that 3^/x = 3 and hence, x — 1. 
Thus, the system of equations above yields the unique solution (1,4). 

Since the constraint defines a closed bounded curve segment, the extreme 
value theorem (Theorem 2.5) applies to guarantee that / must attain both a 
global maximum and a global minimum on this segment. However, the Lagrange 
multiplier method has provided us with just a single critical point. But note that the 
points (9, 0) and (0, 9) satisfy the constraint *Jx + ^/y = 3; they are both points 
where Vg is undefined. Moreover, we have /(l, 4) = 2, while f(9, 0) = 18 and 
f(0, 9) = 9. Evidently then, the minimum of / occurs at (1 , 4) and the maximum 
at (9,0). 

We can understand the geometry of the situation in the following manner. The 
collection of level sets of the function / is the family of parallel lines 2x + y = k. 
Note that the height k of each level set is just the y-intercept of the corresponding 
line in the family. Thus, the problem we are considering is to find the largest and 



4.3 | Lagrange Multipliers 283 



y 




Figure 4.27 The level sets of the function f(x, y) = 2x + y define a family 
of lines. The minimum of / subject to the constraint that *Jx + *Jy = 3 
occurs at a point where one of the lines is tangent to the constraint curve and 
the maximum at one of the endpoints of the curve. 



smallest y -intercepts of any line in the family that meets the curve *Jx + Jy = 3. 
These extreme values of k occur either when one of the lines is tangent to the 
constraint curve or at an endpoint of the curve. (See Figure 4.27.) 

This example illustrates the importance of locating all the points where ex- 
trema may occur by considering places where V/ or Vg is undefined (or where 
Vg = 0) as well as the solutions to the system of equations determined using 
Lagrange multipliers. ♦ 




Figure 4.28 The gradient 
Vg(xo) is perpendicular to 
S = {x | g(x) = c}, hence, to the 
tangent vector at xo to any curve 
x(r) lying in S and passing through 
xo. If / has an extremum at xo, 
then the restriction of / to the 
curve also has an extremum at xq. 



Sketch of a proof of Theorem 3.1 We present the key ideas of the proof, which 
are geometric in nature. Try to visualize the situation for the case n = 3, where 
the constraint equation g(x, y, z) = c defines a surface S inR 3 . (See Figure 4.28.) 
In general, if S is defined as {x | g(x) = c) with Vg(xo) ^ 0, then (at least locally 
near xo) S is a hypersurface in R". The proof that this is the case involves the 
implicit function theorem (Theorem 6.5 in §2.6), and this is why our proof here 
is just a sketch. 

Thus, suppose that x 0 is an extremum of / restricted to S. We consider a 
further restriction of / — to a curve lying in S and passing through x 0 . This will 
enable us to use results from one-variable calculus. The notation and analytic 
particulars are as follows: Let x: / c R —> S C R 3 be a C 1 path lying in S with 
\(to) = xo for some to 6 / . Then the restriction of / to x is given by the function 
F, where 

F(f) = /(x(0). 

Because xo is an extremum of / on S, it must also be an extremum on x. Conse- 
quently, we must have F'(to) = 0, and the chain rule implies that 



= ^/(x(0)| r=ro = V/(x(f 0 )).x'(fo) = V/(x 0 )-x'( f0 ). 



Thus, V/(xo) is perpendicular to any curve in S passing through x 0 ; that is, 
V /(xo) is normal to S at xo. We've seen previously in §2.6 that the gradient 
Vg(xo) is also normal to S at xq. Since the normal direction to the level set S is 



284 Chapter 4 [ Maxima and Minima in Several Variables 



uniquely determined and Vg(xo) 7^ 0, we must conclude that V /(xo) and Vg(xo) 
are parallel vectors. Therefore, 

V/(x 0 ) = AVg(xo) 

for some scalar X e R, as desired. ■ 

The Case of More than One Constraint 

It is natural to generalize the situation of finding extrema of a function / subject 
to a single constraint equation to that of finding extrema subject to several con- 
straints. In other words, we may wish to maximize or minimize / subject to k 
simultaneous conditions of the form 

gi(x) = c\ 

g2(x) = C 2 
g k (x) = C k 

The result that generalizes Theorem 3.1 is as follows: 



THEOREM 3.2 Let X be open in R" and let /, g u . . . , g k : X C R" -> R be C 1 
functions, where k < n.LetS = [x e X \ g\(x) = c\, .. ., gfc(x) = c k }.lf f\ s has 
an extremum at a point xo, where Vgi(xo), . . . , Vgt(xo) are linearly independent 
vectors, then there must exist scalars X\, ... ,X k such that 

V/(xo) = AjVgKxo) + A 2 Vg 2 (x 0 ) + ■ ■ ■ + X k Vg k (x 0 ). 

(Note: k vectors vi , . . . , in R" are said to be linearly independent if the only 
way to satisfy a\ vi + ■ ■ ■ + a k v k = 0 for scalars a\, . . . , a k is if a\ = a% = ■ ■ ■ = 
a k = 0.) 



Idea of proof First, note that S is the intersection of the k hypersurfaces Si , . . . , 
S/c, where Sj = {x 6 R" | g/(x) = cj}. Therefore, any vector tangent to S must 
also be tangent to each of these hypersurfaces, and so, by Theorem 6.4 of 
Chapter 2, perpendicular to each of the Vg ; 's. Given these remarks, the main 
ideas of the proof of Theorem 3.1 can be readily adapted to provide a proof of 
Theorem 3.2. 

Therefore, we let xo e 5 be an extremum of / restricted to S and consider 
the one-variable function obtained by further restricting / to a curve in S through 
xo. Thus, let x: / -> S C R" be a C 1 curve in S with x(f 0 ) = xo for some to £ I. 
Then, as in the proof of Theorem 3.1, we define F by 

F(t) = /(x(0). 

It follows, since x 0 is assumed to be a constrained extremum, that 

F'(t 0 ) = 0. 

The chain rule then tells us that 

0 = F'(to) = V/(x(* 0 )) • x'(to) = V/(xo) • x'(to). 

That is, V /(xo) is perpendicular to all vectors tangent to S at xo. Therefore, it can 
be shown that V/(xq) is in the ^-dimensional plane spanned by the normal vectors 



4.3 | Lagrange Multipliers 285 



Plane spanned by 
Vg^f-^i^o) and 




Figure 4.29 Illustration of the 
proof of Theorem 3.2. The 
constraints gi(x) = c\ and 
g2(x) = C2 are the surfaces S\ and 
S 2 • Any extremum of / must occur 
at points where V/ is in the plane 
spanned by Vgi and Vg 2 . 



to the individual hypersurfaces S\, . .. , S k whose intersection is S. It follows (via 
a little more linear algebra) that there must be scalars ki, . . . , k k such that 



V/(xo) = hVgi(xo) + A 2 Vg 2 (x 0 ) + 



■k k Wg k (x 0 ). 



A suggestion of the geometry of this proof is provided by Figure 4.29 (where 
k = 2 and n = 3). ■ 

EXAMPLE 5 Suppose the cone z 2 = x 2 + y 2 is sliced by the plane z = x + 
y + 2 so that a conic section C is created. We use Lagrange multipliers to find 
the points on C that are nearest to and farthest from the origin in R 3 . 

The problem is to find the minimum and maximum distances from (0, 0, 0) 
of points (x, y, z) on C. For algebraic simplicity, we look at the square of the 
distance rather than the actual distance. Thus, we desire to find the extrema of 

f(x, y, Z ) = x 2 + y 2 + z 2 

(the square of the distance from the origin to (x, y, z)) subject to the constraints 

gi(x, y, z) = x 2 + y 2 - z 2 = 0 
g 2 (x, y, z ) = x + y - z = -2 



Note that 

Vgi(x, y, z) = (2x, 2y, 



-2z) and Vg 2 (x,y,z) = (1,1,-1). 



These vectors are linearly dependent only when x = y = z- However, no point of 
the form (x, x, x) simultaneously satisfies g\ = 0 and g 2 = —2. Hence, Vgi and 
Vg2 are linearly independent at all points that satisfy the two constraints. There- 
fore, by Theorem 3.2, we know that any constrained critical points (x 0 , y 0 , Zo) 
must satisfy 

V/(*o, yo, z 0 ) = kiVgi(x 0 , y 0 , zo) + k 2 Vg 2 (x 0 , yo, zo), 
as well as the two constraint equations. Thus, we must solve the system 

2x = 2k\x + k 2 
2y = 2kiy + k 2 
2z = -2k\z — k 2 . 
x 2 + y 2 -z 2 = 0 
x + y - z = -2 



Eliminating k 2 from the first two equations yields 

k 2 = 2x — 2k\x =2y — 2A.iv, 



which implies that 



Therefore, either 



2(x - y)(l - M) = 0. 



x = y or A.1 = 1. 



The condition ki = 1 implies immediately k 2 = 0, and the third equation of the 
system becomes 2z = —2z, so z must equal 0. If z = 0, then x and y must be 



286 Chapter 4 | Maxima and Minima in Several Variables 

zero by the fourth equation. However, (0, 0, 0) is not a point on the plane z = 
x + y + 2. Thus, the condition X \ = 1 leads to no critical point. On the other hand, 
if x = y, then the constraint equations (the last two in the original system of five) 
become 

2x 2 -z 2 = 0 
2x — z = —2 

Substituting z = 2x + 2 yields 

2x 2 - (2x + 2) 2 = 0, 

equivalent to 

2x 2 + 8x + 4 = 0, 

whose solutions are x = — 2 ± V2- Therefore, there are two constrained critical 
points 

ai = (-2 + 72, -2 + 4l, -2 + 2V2) 

and 

a 2 = (-2 - V2, -2 - V2, -2 - 2V2) . 

We can check that 

/(ai) = 24 - 16 V2, /(a 2 ) = 24 + 16^2, 

so it seems that ai must be the point on C lying nearest the origin, and a 2 must be 
the point that lies farthest. However, we don't know a priori if there is a farthest 
point. If the conic section C is a hyperbola or a parabola, then there is no point 
that is farthest from the origin. To understand what kind of curve C is, note 
that ai has positive z-coordinate and a 2 has negative z-coordinate. Therefore, the 
plane z = x + y + 2 intersects both nappes of the cone z 2 = x 2 + y 2 . The only 
conic section that intersects both nappes of a cone is a hyperbola. Hence, C is a 
hyperbola, and we see that the point ai is indeed the point nearest the origin, but 
the point a 2 is not the farthest point. Instead a 2 is the point nearest the origin on 
the branch of the hyperbola not containing aj. That is, local constrained minima 
occur at both ai and a 2 , but only &\ is the site of the global minimum. (See 
Figure 4.30.) ♦ 

A Hessian Criterion for Constrained Extrema (optional) — 

As Example 5 indicates, it is often possible to determine the nature of a critical 
point (constrained or unconstrained) from considerations particular to the prob- 
lem at hand. Sometimes this is not difficult to do in practice and can provide 
useful insight into the problem. Nonetheless, occasionally it is advantageous to 
have a more automatic means of discerning the nature of a constrained critical 
point. We therefore present a Hessian criterion for constrained critical points. 
Like the one in the unconstrained case, this criterion only determines the local 
nature of a critical point. It does not provide information about global constrained 
extrema. 2 




Figure 4.30 The 

point ai is the point 
on the hyperbola 
closest to the origin. 
The point a 2 is the 
point on the lower 
branch of the 
hyperbola closest 
to the origin. 



2 We invite the reader to consult D. Spring, Amer. Math. Monthly, 92 (1985), no. 9, 631-643 for a more 
complete discussion. 



4.3 | Lagrange Multipliers 287 



In general, the context for the Hessian criterion is this: We seek extrema of a 
function /:XcR"-^R subject to the k constraints 



gi(xi,x 2 , . . .,x„) = a 

g2(X\,X 2 , X„) = C 2 



gk(X\,X2, 



i Xfi) — C k 



We assume that f,gi,...,g k are all of class C 2 , and assume, for simplicity, that 
/ and the gj 's all have the same domain X. Finally, we assume that Vg\ , . . . , Vg k 
are linearly independent at the constrained critical point a. Then, by Theorem 3.2, 
any constrained extremum a must satisfy 

V/(a) = A.i Vgi(a) + A 2 Vg 2 (a) + ■ ■ • + A. t Vg t (a) 

for some scalars Ai , . . . , A*. We can consider a constrained critical point to be a 
pair of vectors 

(A;a) = (Ai, ...,X k ;au ...,a n ) 

satisfying the aforementioned equation. In fact, we can check that (A; a) is an 
unconstrained critical point of the so-called Lagrangian function L defined 
by 



L(h, ...,l k ;xi,...,x„)= f{x\, . . . ,x n ) - y^Ji(gj(xi, ...,x n )- a). 



The Hessian criterion comes from considering the Hessian of L at the critical 
point (A; a). Before we give the criterion, we note the following fact from linear 
algebra: Since Vgi(a), . . . , Vgj(a) are assumed to be linearly independent, the 
derivative matrix of g = (g\ , . . . , g k ) at a, 



Dg(a) = 



dx\ 



(a) 



dx 



(a) 



9gi 
dx n 

dgk 
dx n 



(a) 



(a) 



hasa^: x k submatrix( obtained by deleting n — k columns ofDg(a)) with nonzero 
determinant. By relabeling the variables if necessary, we will assume that 



3jc 



(a) 



dx k 



(a) 



det 



^0 



dx 



(a) 



dx k 



(a) 



(i.e., that we may delete the last n — k columns). 



Chapter 4 I Maxi ma and M 



inima in Several Variables 



Second derivative test for constrained local extrema. Given a constrained 
critical point a of / subject to the conditions g\(x) = c\, g 2 (x) = c 2 , ■ . . , 
g k (x) = Ck, consider the matrix 



HL(X; a) = 



0 



dx 



(a) 



dx n 



(a) 



0 

0 

dgk 
dx\ 

dgk 
dx n 



dx\ 



(a) 



dx n 



(a) 



(a) 



(a) 



dx\ 



(a) 



dx„ 



(a) 



where 



htj = 



d 2 f 



(a) - h - - (a) - X 2 (a) 

OXj dXi 



— —(a). 



dxjdxj dxjdxi dxjdxt dxjdx t 

(Note that HL(k; a) is an (« + /:) x + /c) matrix.) By relabeling the vari- 



ables as necessary, assume that 



det 



3jc 



(a) 



3x 



(a) 



3*<t 



3xi. 



(a) 



(a) 



^0. 



As in the unconstrained case, let be the upper leftmost j x j subma- 
trix of HL(X, a). For j = 1, 2, . . . , k + «, let <2 ; - = det and calculate the 
following sequence of « — k numbers: 

(-1)^+2,..., (1) 

Note that, if k > 1, the sequence in (1) is not the complete sequence of prin- 
cipal minors of HL(X, a). Assume *4 +n = det HL(X, a) / 0. The numerical 
test is as follows: 

1. If the sequence in (1) consists entirely of positive numbers, then / has a 
local minimum at a subject to the constraints 



gi(x) = ci, g 2 (x) = c 2 , 



gk(*) = c k . 



2. If the sequence in (1) begins with a negative number and thereafter alter- 
nates in sign, then / has a local maximum at a subject to the constraints 



gl(x) = Ci, g 2 (x) = C 2 , 



g k (x) = C k . 



3. If neither case 1 nor case 2 holds, then / has a constrained saddle point 
at a. 

In the event that det HL(X, a) = 0, the constrained critical point a is degen- 
erate, and we must use another method to determine whether or not it is the 
site of an extremum. 



4.3 | Lagrange Multipliers 289 



Finally, in the case of no constraint equations g, (x) = c, (i.e., k = 0), the 
preceding criterion becomes the usual Hessian test for a function / of n variables. 

EXAMPLE 6 In Example 1, we found the minimum of the area function 

A(x, v, z) = 2xy + 2yz + xz 
of an open rectangular box subject to the condition 

V(x, y, z) = xyz = 4. 

Using Lagrange multipliers, we found that the only constrained critical point was 
(2, 1 , 2). The value of the multiplier A. corresponding to this point is 2. To use the 
Hessian criterion to check that (2, 1, 2) really does yield a local minimum, we 
construct the Lagrangian function 

L(l;x, y, z) = A(x, y, z) - l(V(x, y, z) - 4) 

= 2xy + 2yz + xz - l{xyz — 4). 



Then 



HL(l;x,y,z) = 



0 

-yz 
-xz 
-xy 



-yz 
0 

2 — Ix 
l-ly 



-xz 
2-lz 
0 

2 — Ix 



-xy 
l-ly 
2 — Ix 
0 



At the constrained critical point (2; 2, 1 , 2), we have 

HL{2;2, 1,2) 

The sequence of determinants to consider is 
(-l) 1 det// 2(1)+1 = 



(-iydet// 4 





0 


-2 


-4 " 






det 


-2 


0 


-2 


= 32, 




-4 


-2 


0 








- 0 


-2 


-4 


-2 " 




det 


-2 
-4 


0 

-2 


-2 
0 


-1 

-2 






_ -2 


-1 


-2 


0 





48. 



Since these numbers are both positive, we see that (2, 1,2) indeed minimizes the 
area of the box subject to the constant volume constraint. ♦ 

EXAMPLE 7 In Example 5, we found points on the conic section C denned 
by equations 

\ gl (x,y,z) = x 2 + y 2 -z 2 = 0 
I gi(x, y, z) = x + y - z = -2 
that are (constrained) critical points of the "distance" function 

f(x,y,z) = x 2 + y 2 + z 2 . 



To apply the Hessian criterion in this case, we construct the Lagrangian function 

z 2 ) - m{x + y - z + 2). 



L(l,m;x, y, z) = x 2 + y 2 + z 2 



l(x 2 + y 2 



290 Chapter 4 I Maxima and Minima in Several Variables 



The critical points of L, found by setting DL(l, m;x,y, z) equal to 0, are 



(A.i;ai) = (-3 + 2V2, 


-24 + 


16^2 ; 


-2 + wl, 


-2 + wl, - 


-2 + 


and 












(A 2 ;a 2 ) = (-3-2V2,- 


-24- 


16V2;- 


-2 - y/2, 


-2-V2, - 


- 2 - 


The Hessian of L is 














0 


0 


—2x 


-2y 


2z 




0 


0 


-1 


-1 


1 


HL(l, m;x, y, z) = 


— 2x 


-1 


2-2/ 


0 


0 




-2y 


-1 


0 


2-2/ 


0 




2z 


1 


0 


0 2 


+ 21 



2V2). 



After we evaluate this matrix at each of the critical points, we need to compute 

(-l) 2 det// 2(2)+1 =det// 5 . 

We leave it to you to check that for (k\; ai) this determinant is 128 — 64 V2 f*s 
37.49, and for (A 2 ; a 2 ) it is 128 + 64V2 & 218.51. Since both numbers are pos- 
itive, the points (—2 ± V2> —2 ± -Jl, —2 ± 2%/2) are both sites of local min- 
ima. By comparing the values of / at these two points, we see that (—2 + \/2, 
—2 + V2, —2 + 2*Jl) must be the global minimum. ♦ 



4.3 Exercises 



1 . In this problem, find the point on the plane 2x — 3 y — 
z = 4 that is closest to the origin in two ways: 

(a) by using the methods in §4.2 (i.e., by finding the 
minimum value of an appropriate function of two 
variables); 

(b) by using a Lagrange multiplier. 

In Exercises 2-12, use Lagrange multipliers to identify the crit- 
ical points of f subject to the given constraints. 

2. f(x, y) = y, 2x 2 + y 2 = 4 

3. f(x, y) = 5x + 2y, 5x 2 + 2y 2 = 14 

4. f(x, y) = xy, 2x — 3y = 6 

5. f(x, y, z) = xyz, 2x + 3 y + z = 6 

6. f(x, y,z) = x 2 + y 2 + z 2 , X + y - z = 1 

7. f(x,y,z) = 3-x 2 -2y 2 -z 2 , 2x + y + z = 2 

8. f(x, y, z) = x 6 + y 6 + z 6 , x 2 + y 2 + z 2 = 6 

9. f(x, y, z) = 2x + y 2 -z 2 , x - 2y = 0, x + z = 0 

10. f(x, y, z) = 2x + y 2 + 2z, x 2 - y 2 = 1, x + y + 
z = 2 

11. f(x, y, z) = xy + yz, x 2 + y 2 = 1, yz = 1 



12. f(x, y,z) = x + y + z, y 2 - x 2 = 1, x + 2z=l 

13. (a) Find the critical points of f(x, y) = x 2 + y sub- 

ject to x 2 + 2y 2 = 1. 

(b) Use the Hessian criterion to determine the nature 
of the critical point. 

14. (a) Find any critical points of f(x, y, z, w) = x 2 + 

y 2 + z 2 + w 2 subject to 2x + y + z = 1 , x — 2z — 
w = —2, 3x + y + 2w = — 1. 

(b) Use the Hessian criterion to determine the nature 
of the critical point. (Note: You may wish to use a 
computer algebra system for the calculations.) 

Just as sometimes is the case when finding ordinary (i.e., un- 
constrained) critical points of functions, it can be difficult to 
solve a Lagrange multiplier problem because the system of 
equations that results may be prohibitively difficult to solve by 
hand. In Exercises 15—19, use a computer algebra system to 
find the critical points of the given function f subject to the 
constraints indicated. (Note: You may find it helpful to provide 
numerical approximations in some cases.) 

^ 15. f(x, y, z) = 3xy - 4z, 3x + y - 2xz = 1 

^ 16. f(x, y, z) = 3xy - 4yz + 5xz, 3x + y + 2z= 12, 

2x - 3y + 5z = 0 



4.3 I Exercises 291 



17. f(x, y, z) = y 3 + 2xyz - x 2 , x 2 + y 2 + z 2 = 1 

18. f(x, y, z) = x 2 + y 2 - xz 2 , xy + z 2 = 1 

19. f(x, y, z, w) = x 2 + y 2 + z 2 + w 2 , x 2 + y 2 = 1, 
x + y + z + w = \,x — y + z — w = 0 

20. Consider the problem of determining the extreme val- 
ues of the function f(x, y) = x 3 + 3y 2 subject to the 
constraint that xy = —A. 

(a) Use a Lagrange multiplier to find the critical points 
of / that satisfy the constraint. 

(b) Give an analytic argument to determine if the criti- 
cal points you found in part (a) yield (constrained) 
maxima or minima of /. 

(c) Use a computer to plot, on a single set of axes, sev- 
eral level curves of / together with the constraint 
curve xy = —A. Use your plot to give a geometric 
justification for your answers in parts (a) and (b). 

21. Find three positive numbers whose sum is 18 and 
whose product is as large as possible. 

22. Find the maximum and minimum values of 
f(x, y, z) = x + y — z on the sphere x 2 + y 2 + z 2 = 
8 1 . Explain how you know that there must be both a 
maximum and a minimum attained. 

23. Find the maximum and minimum values of f(x, y) = 
x 2 + xy + v 2 on the closed disk D = {(x, y) | x 2 + y 2 
<4}. ' 

24. You are sending a birthday present to your calculus in- 
structor. Fly-By-Night Delivery Service insists that any 
package it ships be such that the sum of the length plus 
the girth be at most 1 08 in. (The girth is the perimeter of 
the cross section perpendicular to the length axis — see 
Figure 4.31.) What are the dimensions of the largest 
present you can send? 




Figure 4.31 Diagram for 
Exercise 24. 



25. A cylindrical metal can is to be manufactured from 
a fixed amount of sheet metal. Use the method of 
Lagrange multipliers to determine the ratio between 
the dimensions of the can with the largest capacity. 



26. An industrious farmer is designing a silo to hold her 
900tt ft 3 supply of grain. The silo is to be cylindrical 
in shape with a hemispherical roof. (See Figure 4.32.) 
Suppose that it costs five times as much (per square 
foot of sheet metal used) to fashion the roof of the silo 
as it does to make the circular floor and twice as much 
to make the cylindrical walls as the floor. If you were 
to act as consultant for this project, what dimensions 
would you recommend so that the total cost would be 
a minimum? On what do you base your recommen- 
dation? (Assume that the entire silo can be filled with 
grain.) 




Figure 4.32 The grain silo 
of Exercise 26. 

27. You are in charge of erecting a space probe on the 
newly discovered planet Nilrebo. To minimize interfer- 
ence to the probe's sensors, you must place the probe 
where the magnetic field of the planet is weakest. Nil- 
rebo is perfectly spherical with a radius of 3 (where 
the units are thousands of miles). Based on a coordi- 
nate system whose origin is at the center of Nilrebo, 
the strength of the magnetic field in space is given by 
the function M(x, y, z) = xz — y 2 + 3x + 3. Where 
should you locate the probe? 

28. Heron's formula for the area of a triangle whose sides 
have lengths x, y, and z is 

Area = ^s(s — x)(s — y)(s — z), 

where s = j(x + y + z) is the so-called semi- 
perimeter of the triangle. Use Heron's formula to show 
that, for a fixed perimeter P, the triangle with the 
largest area is equilateral. 

29. Use a Lagrange multiplier to find the largest sphere 
centered at the origin that can be inscribed in the ellip- 
soid 3x 2 + 2y 2 + z 2 = 6. (Be careful with this prob- 
lem; drawing a picture may help.) 

30. Find the point closest to the origin and on the line 
of intersection of the planes 2x + y + 3z = 9 and 
3x + 2y + z = 6. 



292 Chapter 4 I Maxima and Minima in Several Variables 



31 . Find the point closest to the point (2,5,-1) and on the 
line of intersection of the planes x — 2y + 3z = 8 and 

2z-y = 3. 

32. The plane x + y + z = 4 intersects the paraboloid 
z = x 2 + y 2 in an ellipse. Find the points on the el- 
lipse nearest to and farthest from the origin. 

33. Find the highest and lowest points on the ellipse ob- 



tained by intersecting the paraboloid z 
the plane x + y + 1z = 2, 



+ y z with 



34. Find the minimum distance between a point on the 
ellipse x 2 + 2y 2 = 1 and a point on the line x + y = 4. 
(Hint: Consider a point (x , y ) on the ellipse and a point 
(u, v) on the line. Minimize the square of the distance 
between them as a function of four variables. This prob- 
lem is difficult to solve without a computer.) 

35. (a) Use the method of Lagrange multipliers to find crit- 

ical points of the function f(x, y) = x + y subject 
to the constraint xy = 6. 

(b) Explain geometrically why / has no extrema on 
the set {(x, y) \ xy = 6}. 

36. Leta, fi, and y denote the (interior) angles ofatriangle. 
Determine the maximum value of sin a sin /3 sin y . 

37. Let S be a surface in R 3 given by the equation 
g(x, y, z) = c, where g is a function of class C 1 with 
nonvanishing gradient and c is a constant. Suppose that 
there is a point P on S whose distance from the origin 
is a maximum. Show that the displacement vector from 
the origin to P must be perpendicular to S. 

38. The cylinder x 2 + y 2 = 4 and the plane 2x + 2y + 
z = 2 intersect in an ellipse. Find the points on 
the ellipse that are nearest to and farthest from the 
origin. 

39. Find the points on the ellipse 3x 2 — Axy + 3y 2 = 50 
that are nearest to and farthest from the origin. 

40. This problem concerns the determination of the ex- 
trema of f(x, y) = ~Jx + subject to the con- 
straint x 2 + y 2 = 17, where x > 0 and y > 0. 

(a) Explain why / must attain both a global minimum 
and a global maximum on the given constraint 
curve. 

(b) Use a Lagrange multiplier to solve the system of 
equations 

\Vf(x,y) = XVg{x,y) 

U(x, y) = o 

where g(x, y) = x 2 + y 2 . You should identify a 
single critical point of /. 

(c) Identify the global minimum and the global max- 
imum of / subject to the constraint. 



41 . Consider the problem of finding extrema of f(x, y) = 
x subject to the constraint y 2 — 4x 3 + 4x 4 = 0. 

(a) Use a Lagrange multiplier and solve the system of 
equations 

Wf(x,y) = Wg(x,y) 
[g(x,y) = 0 

where g(x, y) = y 2 — 4x 3 + 4x A . By doing so, 
you will identify critical points of / subject to the 
given constraint. 

(b) Graph the curve y 2 — 4x 3 + 4x 4 = 0 and use the 
graph to determine where the extrema of f(x, y) = 
x occur. 

(c) Compare your result in part (a) with what you 
found in part (b). What accounts for any differ- 
ences that you observed? 

42. Consider the problem of finding extrema of 
f(x, y, z) = x 2 + y 2 subject to the constraint z = c, 
where c is any constant. 

(a) Use the method of Lagrange multipliers to iden- 
tify the critical points of / subject to the constraint 
given above. 

(b) Using the usual alphabetical ordering of variables 
(i.e., x\ =x,x2 = y, x^ = z), construct the Hessian 
matrix HL(X;a\, 02, 03) (where L(l;x, y, z) = 
f{x, y, z) — l(z — c)) for each critical point you 
found in part (a). Try to use the second deriva- 
tive test for constrained extrema to determine the 
nature of the critical points you found in part (a). 
What happens? 

(c) Repeat part (b), this time using the variable order- 
ing x\ = z, X2 = y, X3 = x. What does the second 
derivative test tell you now? 

(d) Without making any detailed calculations, discuss 
why / must attain its minimum value at the point 
(0, 0, c). Then try to reconcile your results in parts 
(b) and (c). This exercise demonstrates that the as- 
sumption that 



det 



dx 



dx 



(a) 



(a) 



dx k 



dx k 



(a) 



(a) 



#0 



is important. 



43. Consider the problem of finding critical points of the 
function f(xi, ■ ■ ■ , x„) subject to the set of k con- 
straints 

gl(Xl, ...,x„) = ci, gi{x u ...,*„) = c 2 , ... , 
gk(x\, ...,x„) = c k . 
Assume that f, gi, g2, • ■ ■ , gk are all of class C 2 . 



4.4 | Some Applications of Extrema 



(a) Show that we can relate the method of Lagrange 
multipliers for determining constrained critical 
points to the techniques in §4.2 for finding un- 
constrained critical points as follows: If 

(k, a) = (ki, . . . , k k ;a u . . . , a n ) 

is a pair consisting of k values for Lagrange mul- 
tipliers k\, . . . , kk and n values a\ , . . . , a„ for the 
variables X\, .. . ,x„ such that a is a constrained 
critical point, then (k, a) is an ordinary (i.e., un- 
constrained) critical point of the function 

LQi, ...,l k ;xi, ...,x„) 

k 

= f(X\, ...,X n )~ } li(gi(Xl, ... ,X n )- Ci). 

1=1 

(b) Calculate the Hessian HL(k, a), and verify that it 
is the matrix used in §4.3 to provide the criterion 
for determining the nature of constrained critical 
points. 



44. The unit hypersphere in R" (centered at the origin 
0 = (0, . . . , 0)) is defined by the equation xj+xj + 

■ ■ ■ + x\ = 1. Find the pair of points x = (xi x n ) 

and y = (vi, . • • , y„), each of which lies on the unit 
hypersphere, that maximizes and minimizes the func- 
tion 

n 

f(x\, X n , Vi, . . • , y„) = xtyt. 

i=l 

What are the maximum and minimum values of /? 

45. Let x = (xi, . . . , Xn) and y = (y\, . . . , y n ) be any vec- 
tors in R" and, for = 1, . . . , n, set 

u; = = and v, = =. 

(a) Show that u = («i , . . . , u n ) and v = (i>i , . . . , v n ) 
lie on the unit hypersphere in R" . 

(b) Use the result of Exercise 44 to establish the 
Cauchy-Schwarz inequality 

|x-y| < II x || ||y||. 



4.4 Some Applications of Extrema 

In this section, we present several applications of the methods for finding both 
constrained and unconstrained extrema discussed previously. 



53 
X 



Protein level 



Figure 4.33 Height versus protein 
level. 




Protein level 



Figure 4.34 Fitting a line to 
the data. 



Least Squares Approximation 

The simplest relation between two quantities x and y is, without doubt, a linear 
one: y = mx + b (where m and b are constants). When a biologist, chemist, 
psychologist, or economist postulates the most direct connection between two 
types of observed data, that connection is assumed to be linear. Suppose that Bob 
Biologist and Carol Chemist have measured certain blood protein levels in an 
adult population and have graphed these levels versus the heights of the subjects 
as in Figure 4.33. If Prof. Biologist and Dr. Chemist assume a linear relationship 
between the protein and height, then they desire to pass a line through the data as 
closely as possible, as suggested by Figure 4.34. 

To make this standard empirical method of linear regression precise (in- 
stead of merely graphical and intuitive), we first need some notation. Suppose we 
have collected n pairs of data (x\ , yi), (x2, yi), • ■ ■ , (x n , y„ ). (In the example just 
described, x t is the protein level of the ith subject and y, his or her height.) We 
assume that there is some underlying relationship of the form y = mx + b, and 
we want to find the constants m and b so that the line fits the data as accurately 
as possible. Normally, we use the method of least squares. The idea is to find 
the values of m and b that minimize the sum of the squares of the differences 
between the observed y-values and those predicted by the linear formula. That is, 
we minimize the quantity 



D(m, b) = [vi - {mx\ + b)f + [y 2 - (mx 2 + b)f 
+ --- + [y n -(mx n +b)f, 



(1) 



294 Chapter 4 | Maxima and Minima in Several Variables 



Line y = mx + b 




Figure 4.35 The method of least squares. 

where, for i = 1, . . . , n, y,- represents the observed y-value of the data, and 
mxi + b represents the y-value predicted by the linear relationship. Hence, each 
expression in D of the form y, — (mx, + b) represents the error between the 
observed and predicted y-values. (See Figure 4.35.) They are squared in the 
expression for D in order to avoid the possibility of having large negative 
and positive terms cancel one another, thereby leaving little or no "net error," 
which would be misleading. Moreover, D(m , b) is the square of the distance 
in R" between the point (yi, y2, . . . , y„) and the point {mx\ + b, mx 2 + b, . . . , 
mx n + b). 

Thus, we have an ordinary minimization problem at hand. To solve it, we 
need to find the critical points of D. First, we can rewrite D as 

n 

D(m, b) = y^[yj - (mxi + b)] 2 

n n n n 

= ^2 yf - 2m X! Xtyi ~ 2b ^2, yi + XI ( mxi + fo ) 2 - 
i=i i=i i=i 

H ii 

1=1 1=1 

ii n n 

2 ^2 x i Vi + 2m xf + 2b x\ 

i=l i=l 1 = 1 

ii n 

-- -2 yt + X! 2{,nXi + b "> 

1 = 1 i=l 

11 17 

= —2 y,- + 2m X; + 2nb. 

i = l i=l 

When we set both partial derivatives equal to zero, we obtain the following pair 
of equations, which have been simplified slightly: 

{H x ?) m + (H x i) b = H x >yi 

(2) 

(J2xj)m + nb = J2yi 

(All sums are taken from i = 1 to «.) Although (2) may look complicated, it is 
nothing more than a linear system of two equations in the two unknowns m and b. 



Then 



i=i 



dD 

dm 



and 



3D 

~db 



4.4 | Some Applications of Extrema 



It is not difficult to see that system (2) has a single solution. Therefore, we have 
shown the following: 



PROPOSITION 4.1 Given n data points (jci, yi), (x 2 , yi), . . . , (x„, y„) with not 
all of x\, X2, . . ■ , x n equal, the function 

n 

D(m,b) = J2[y l -(mx l +b)f 
i=i 

has a single critical point (mo, bo) given by 

mo = j ' 

and 

C>0 = 2 " 



Since D(m , b) is a quadratic polynomial in m and b, the graph of z = D(m,b) 
is a quadric surface. (See §2. 1 .) The only such surfaces that are graphs of functions 
are paraboloids and hyperbolic paraboloids. We show that, in the present case, 
the graph is that of a paraboloid by demonstrating that D has a local minimum at 
the critical point (mo, &o) given in Proposition 4.1. 

We can use the Hessian criterion to check that D has a local minimum at 
(mo, bo). We have 



HD(m, b) = 



2 ^2 Xj 2n 



The principal minors are and An J2 x f ~ 4(X>,) • The first minor is 

obviously positive, but determining the sign of the second requires a bit more 
algebra. (If you wish, you can omit reading the details of this next calculation 
and rest assured that the story has a happy ending.) Ignoring the factor of 4, we 

examine the expression n xf — (J2 *;) 2 - Expanding the second term yields 



»£*?-(£■ 



= n J2 x > - [ E x ' 2 + J2 2x ' x j ) 
i=i \ i=i i<j ) 

n 

= {n - l)^ x ' 2 ~ ^2 2x i x J- 

i'=l i<j 

On the other hand we have 

n 

y^x x i ~ x j^ 2 = £ ( x f ~ 2x ' x j + x )) = (« - 1) £ x f - £ 2xjXj. 

(To see that equation (4) holds, you need to convince yourself that 

£tf + *5) = 0i-i)I>? 



(3) 



(4) 



296 Chapter 4 | Maxima and Minima in Several Variables 



by counting the number of times a particular term of the form x\ appears in the 
left-hand sum.) Thus, we have 

det HD(m,b) = 4 I n^x 2 - f^*, 



Best fit 
line 



* (5, 4) 




Figure 4.36 Data for the linear 
regression of Example 1. 



= 4 1 (n — 1) x 2 — IxiXj j by equation (3), 
\ i=i <</' / 

= 4 / ,(x,- — Xj) 2 by equation (4). 



KJ 

Because this last expression is a sum of squares, it is nonnegative. Therefore, the 
Hessian criterion shows that D does indeed have a local minimum at the critical 
point. Hence, the graph of z = D(m, b) is that of a paraboloid. Since the (unique) 
local minimum of a paraboloid is in fact a global minimum (consider a typical 
graph), we see that D is indeed minimized at (m 0 , b 0 ). 

EXAMPLE 1 To see how the preceding discussion applies to a specific set of 
data, consider the situation depicted in Figure 4.36. 

We have n = 5, and the function D to be minimized is 

D(m, b) = [2- (m + b)f + [1 - (2m + b)f + [5 - (3m + b)] 2 

+ [3 - (4m + b)f + [4 - (5m + b)f. 



We compute 

= 15, 

Thus, using Proposition 4.1, 

5-51-15-15 



m 



5-55-15-15 



55, 

_ 3 

" 5' 



15, 



55 ■ 15 - 15 ■ 51 _ 6 
5-55-15-15 ~ 5' 



The best fit line in terms of least squares approximation is 

3 6 
V = -x H — . 
"^5 5 



Of course, linear regression is not always an appropriate technique. It may 
not be reasonable to assume that the data points fall nearly on a straight line. 
Some formula other than y = mx + b may have to be assumed to describe the 
data with any accuracy. Such a postulated relation might be quadratic, 

y = ax 2 + bx + c, 

or x and y might be inversely related 

a 

y=-+b. 

x 

You can still apply the method of least squares to construct a function analogous 
to D in equation (1) to find the relation of a given form that best fits the data. 

Another way that least squares arise is if y depends not on one variable but 
on several: x\, X2, ■ ■ ■ , x n . For example, perhaps adult height is measured against 



4.4 | Some Applications of Extrema 




Figure 4.37 A particle traveling 
in a force field F. 



blood levels of 10 different proteins instead of just one. Multiple regression is 
the statistical method of finding the linear function 



y = a\X\ + a 2 x 2 + 
that best fits a data set of (n + l)-tuples 



+ a„x n + b 



(a- 



(i) 



x 1 



,xi l \ yi),(*r.*2 



(2) (2) 



{x 



(*) V W 
1 ' x 2 ' 



*,fU) 



We can find such a "best fit hyperplane" by minimizing the sum of the squares of 
the differences between the y-values furnished by the data set and those predicted 
by the linear formula. We leave the details to you. 3 

Physical Equilibria 



Let F: X c R 3 — > R 3 be a continuous force field acting on a particle that moves 
along a path x: / c R —> R 3 as in Figure 4.37. Newton's second law of motion 
states that 



F(x(f)) = mx"(t), 



(5) 



where m is the mass of the particle. For the remainder of this discussion, we will 
assume that F is a gradient field, that is, that F = — VV for some C 1 potential 
function V:X c R 3 -> R. (See §3.3 for a brief comment about the negative sign.) 
We first establish the law of conservation of energy. 



THEOREM 4.2 (Conservation of energy) Given the set-up above, the 
quantity 



\m\\x!{t)\\ 2 + V(x(0) 



is constant. 



The term hn ||x'(?) || 2 is usually referred to as the kinetic energy of the particle 
and the term V(x(r)) as the potential energy. The significance of Theorem 4.2 
is that it states that the sum of the kinetic and potential energies of a particle 
is always fixed (conserved) when the particle travels along a path in a gradient 
vector field. For this reason, gradient vector fields are also called conservative 
vector fields. 

Proof of Theorem 4.2 As usual, we show that the total energy is constant by 
showing that its derivative is zero. Thus, using the product rule and the chain rule, 
we calculate 

d r, 



dt L 



i'(0 • x'(0 + V(x(/))] = mx"(t) ■ x'(t) + V V(x(0) • x'(0 

= mx"(t) • x'(t) - F(x(0) • x'(t) 
= mx"(t) • x'(t) - mx"(t) • x'(t) 
= 0, 

from the definitions of F and V and by formula (5). 



3 Or you might consult S.Weisberg, Applied Linear Regression, 2nd ed., Wiley-Interscience, 1985, 
Chapter 2. Be forewarned, however, that to treat multiple regression with any elegance requires somewhat 
more linear algebra than we have presented. 



298 Chapter 4 | Maxima and Minima in Several Variables 




Figure 4.38 For a stable 
equilibrium point, the path of a 
nearby particle with a sufficiently 
small kinetic energy will remain 
nearby with a bounded kinetic 
energy. 



In physical applications it is important to identify those points in space that 
are "rest positions" for particles moving under the influence of a force field. These 
positions, known as equilibrium points, are such that the force field does not act 
on the particle so as to move it from that position. Equilibrium points are of two 
kinds: stable equilibria, namely, equilibrium points such that a particle perturbed 
slightly from these positions tends to remain nearby (for example, a pendulum 
hanging down at rest) and unstable equilibria, such as the act of balancing a ball 
on your nose. The precise definition is somewhat technical. 



DEFINITION 4.3 Let F: X c R" -> R" be any force field. Then x 0 e X is 
called an equilibrium point of F if F(xo) = 0. An equilibrium point xo is 
said to be stable if, for every r, e > 0, we can find other numbers ro, eo > 0 
such that if we place a particle at position x with || x — x 0 1| < r 0 and provide it 
with a kinetic energy less than eo, then the particle will always remain within 
distance r of xq with kinetic energy less than e. 



In other words, a stable equilibrium point xo has the following property: You 
can keep a particle inside a specific ball centered at xo with a small kinetic energy 
by starting the particle inside some other (possibly smaller) ball about x 0 and 
imparting to it some (possibly smaller) initial kinetic energy. (See Figure 4.38.) 

THEOREM 4.4 For a C 1 potential function V of a vector field F = -V V, 

1. The critical points of the potential function are precisely the equilibrium 
points of F. 

2. If x 0 gives a strict local minimum of V, then x 0 is a stable equilibrium point 
ofF. 



EXAMPLE 2 The vector field F = (-6x - 2y - 2)i + {-2x - Ay + 2)j is 
conservative and has 

V(x, y) = 3x 2 + 2xy + 2x + 2y 2 - 2y + 4 

as a potential function (meaning that F = — V V, according to our current sign 
convention). There is only one equilibrium point, namely, (— | , |) . To see if it is 
stable, we look at the Hessian of V : 



6 2 
2 4 



The sequence of principal minors is 6, 20. By the Hessian criterion, (— |, |) is a 
strict local minimum of V and, by Theorem 4.4, it must be a stable equilibrium 
point of F. ♦ 

Proof of Theorem 4.4 The proof of part 1 is straightforward. Since F = — VV, 
we see that F(x) = 0 if and only if W(x) = 0. Thus, equilibrium points of F are 
the critical points of V. 

To prove part 2, let xo be a strict local minimum of V and x:/^R"aC' 
path such that x(? 0 ) = x 0 for some to e /. By conservation of energy, we must 
have, for all t 6 /, that 

im||x'(0H 2 + V(x(/)) = jm\\x'(t 0 )\\ 2 + V(x(t 0 )). 



4.4 | Some Applications of Extrema 




Figure 4.39 On the surface 
S = {x ] g{\) = c], the component 
of F that is tangent to S at x is 
denoted by $(x). 



To show that xo is a stable equilibrium point, we desire to show that we can bound 
the distance between x(/) and xo = x(fo) by any amount r and the kinetic energy 
by any amount e. That is, we want to show we can achieve 

||x(O-x 0 || < r 

(i.e., x(?) e B r (x 0 ) in the notation of §2.2) and 

im||x'(OH 2 < e. 

As the particle moves along x away from xo, the potential energy must increase 
(since xo is assumed to be a strict local minimum of potential energy), so the 
kinetic energy must decrease by the same amount. For the particle to escape from 
B, (x 0 ), the potential energy must increase by a certain amount. If eo is chosen to 
be smaller than that amount, then the kinetic energy cannot decrease sufficiently 
(so that the conservation equation holds) without becoming negative. This being 
clearly impossible, the particle cannot escape from B r (xo). ■ 

Often a particle is not only acted on by a force field but also constrained to lie 
in a surface in space. The set-up is as follows: F is a continuous vector field on R 3 
acting on a particle that lies in the surface S = {x 6 R 3 | g(x) = c}, where g is a 
C 1 function such that Vg(x) ^ 0 for all x in S. Most of the comments made in the 
unconstrained case still hold true, provided F is replaced by the vector component 
of F tangent to S. Since, at x e S, Vg(x) is normal to S, this tangential component 
of F at x is 

<Kx) = F(x) - proj Vi? (x)F(x). (6) 
(SeeFigure4.39.)Theninplaceofformula(5),wehave,forapathx: / cR->S, 

$(x(f)) = mx"(t). (7) 
We can now state a "constrained version" of Theorem 4.4. 



THEOREM 4.5 For a C 1 potential function V of a vector field F = - V V, 

1. If V|s has an extremum at xo e 5, then xo is an equilibrium point in S. 

2. If V | s has a strict local minimum at x 0 e S, then x 0 is a stable equilibrium 
point. 



Sketch of proof Forpart 1, if V\s has an extremum at xo, then, by Theorem 3.1, 
we have, for some scalar X, that 

VV(x 0 ) = AVg(x 0 ). 

Hence, because F = — VV, 

F(x 0 ) = -AVg(xo), 

implying that F is normal to 5 at xo. Thus, there can be no component of F tangent 
to S at xo (i.e., 4>(xo) = 0). Since the particle is constrained to lie in S, we see 
that the particle is in equilibrium in S. 

The proof of part 2 is essentially the same as the proof of part 2 of Theorem 
4.4. The main modification is that the conservation of energy formula in Theorem 
4.2 must be established anew, as its derivation rests on formula (5), which has 
been replaced by formula (7). Consequently, using the product and chain rules, 



Chapter 4 i Maxima and Minima in Several Variables 



x 2 + y 2 + (z - 2r) 2 





(0, 0, 3r) 


= r 2 -i 




f F ' 






'(0,0./-) 



Figure 4.40 On the sphere 
x 2 + y 2 + (z- 2r) 2 = r 2 , the 
points (0, 0, r) and (0, 0, 3r) 
are equilibrium points for the 
gravitational force field 
F = — mgk. 



we check, for x: / 
d 
dt 



m\\x'(t)\\ 2 + V(x(r))] = j t [^mx'(t).x'(t)+ V(x(t))] 



= mx"(t) • x'(t) + VV(x(/)) • x'(0- 



Then, using formula (6), we have 
d 



dt 



m \\x'(t)\\ 2 + V(x(0)] = x'(t) • mx"(t) - F(x(f )) • x'(t) 

= x'(t).<t>(x(t))-F(x(t)).x'(t) 

= x'(t) ■ [F(x(0) - proj Vg( x(r))F(x(0)] 

-F(x(/))-x'(f) 
= -x'(0-proj Vs (x(r))F(x(0) 



after cancellation. Thus, we conclude that 
d 
dt 

since x'(t ) is tangent to the path in S and, hence, tangent to S itself at x(t), while 
projvo( X (j))F(x(0) is parallel to Vg(x(f )) and, hence, perpendicular to S at x(r). ■ 



m||x'(OII 2 + V(x(0)] = 0, 



EXAMPLE 3 Near the surface of the earth, the gravitational field is ap- 
proximately 

F = —mgk. 

(We're assuming that, locally, the surface of the earth is represented by the plane 
z = 0.) Note that F = -V V, where 

V(x, y, z) = mgz. 

Now suppose a particle of mass m lies on a small sphere with equation 

h(x, y, z ) = x 2 + y 2 + (z- 2rf = r 2 . 

We can find constrained equilibria for this situation, using a Lagrange multiplier. 
The gradient equation V V = XV h, along with the constraint, yields the system 

0 = 2Ax 
0 = 2ky 

mg = 2X(z — 2r) 
x 2 + y 2 + (z - 2r) 2 = r 2 

Because m and g are nonzero, X cannot be zero. The first two equations imply 
x = y = 0. Therefore, the last equation becomes 



which implies 



( z -2r) 2 = r 2 , 



Z = r, 3r 



are the solutions. Consequently, the positions of equilibrium are (0, 0, r) and 
(0, 0, 3r) (corresponding to X = —mg/2r and +mg/2r, respectively). From ge- 
ometric considerations, we see V is strictly minimized at S at (0, 0, r) and maxi- 
mized at (0, 0, 3r) as shown in Figure 4.40. From physical considerations, (0, 0, r) 



4.4 | Some Applications of Extrema 



is a stable equilibrium and (0, 0, 3r) is an unstable one. (Try balancing a marble 



Applications to Economics 

We present two illustrations of how Lagrange multipliers occur in problems in- 
volving economic models. 

EXAMPLE 4 The usefulness of amounts x\, x%, . . . , x n of (respectively) dif- 
ferent capital goods G\,Gi, ...,G n can sometimes be measured by a function 
U(x\, X2, ■ . . , x n ), called the utility of these goods. Perhaps the goods are indi- 
vidual electronic components needed in the manufacture of a stereo or computer, 
or perhaps U measures an individual consumer's utility for different commodi- 
ties available at different prices. If item G, costs a, per unit and if M is the total 
amount of money allocated for the purchase of these n goods, then the consumer 
or the company needs to maximize U(x\, x% x„) subject to 



This is a standard constrained optimization problem that can readily be approached 
by using the method of Lagrange multipliers. 

For instance, suppose you have a job ordering stationery supplies for an office. 
The office needs three different types of products a, b, and c, which you will order 
in amounts x, y, and z, respectively. The usefulness of these products to the smooth 
operation of the office turns out to be modeled fairly well by the utility function 
U(x, y, z) = xy + xyz. If product a costs $3 per unit, product b $2 a unit, and 
product c $ 1 a unit and the budget allows a total expenditure of not more than 
$899, what should you do? The answer should be clear: You need to maximize 

U(x, y, z) = xy + xyz subject to B(x, y, z) = 3x + 2y + z = 899. 

The Lagrange multiplier equation, VC/(x, y, z) = XVB(x, y, z), and the budget 
constraint yield the system 



The last equality implies that either x = 0 or y = (z + l)/2. We can reject the 
first possibility, since t/(0, y, z) = 0 and the utility U(x, y, z) > 0 whenever 
x, y, and z are all positive. Thus, we are left with y = (z + l)/2. This in turn 
implies that A. = (z + l) 2 /6. Substituting for y in the constraint equation shows 
that x = (898 — 2z)/3, so that equation xy = X becomes 



on top of a ball.) 



♦ 



a^xi + 02X2 + ■ ■ ■ + a„x n = M. 



y + yz = 3A 
x + xz = 2k 
xy = X 

3x + 2y + z = 899 



Solving for X in the first three equations yields 





which is satisfied by either z = — 1 (which we reject) or by z = 299. The only 
realistic critical point for this problem is (100, 150, 299). We leave it to you to 
check that this point is indeed the site of a maximum value for the utility. ♦ 



Chapter 4 i Maxima and Minima in Several Variables 



EXAMPLE 5 In 1928, C.W.Cobb and P. M.Douglas developed a simple model 
for the gross output Q of a company or a nation, indicated by the function 

Q(K,L) = AK"L 1 -", 

where K represents the capital investment (in the form of machinery or other 
equipment), L the amount of labor used, and A and a positive constants with 
0 < a < 1 . (The function Q is known now as the Cobb-Douglas production 
function.) If you are president of a company or nation, you naturally wish to 
maximize output, but equipment and labor cost money and you have a total 
amount of M dollars to invest. If the price of capital is p dollars per unit and 
the cost of labor (in the form of wages) is w dollars per unit, so that you are 
constrained by 

B(K, L) = pK + wL < M, 

what do you do? 

Again, we have a situation ripe for the use of Lagrange multipliers. Before 
we consider the technical formalities, however, we consider a graphical solution. 
Draw the level curves of Q, called isoquants, as in Figure 4.41. Note that Q 
increases as we move away from the origin in the first quadrant. The budget 
constraint means that you can only consider values of K and L that lie inside or 
on the shaded triangle. It is clear that the optimum solution occurs at the point 
(K, L) where the level curve is tangent to the constraint line pK + wL = M. 

Here is the analytical solution: From the equation VQ(K, L) = XVB(K, L) 
plus the constraint, we obtain the system 

'AaK"- l L l - a = Xp 
A(l - d)K a lr a = Xw. 
pK + wL = M 



Solving for p and w in the first two equations yields 




K 



Figure 4.41 A family of isoquants. The optimum value of Q(K, L) subject to the 
constraint pK + wL = M occurs where a curve of the form Q = c is tangent to the 
constraint line. 



4.4 I Exercises 303 



Substitution of these values into the third equation gives 

Aa „ i „ A(l - a) „ , „ 

— K L + — -K"L l - a = M. 

X X 

Thus, 

X = —K a L 1 -", 
M 

and the only critical point is 

(Ma M(\-a) 
\ P w 

From this geometric discussion, we know that the critical point must yield the 
maximum output Q. 

From the Lagrange multiplier equation, at the optimum values for L and K, 
we have 

p dK w dL 

This relation says that, at the optimum values, the marginal change in output per 
dollar's worth of extra capital equals the marginal change per dollar's worth of 
extra labor. In other words, at the optimum values, exchanging labor for capital 
(or vice versa) won't change the output. This is by no means the case away from 
the optimum values. 

There is not much that is special about the function Q chosen. Most of our 
observations remain true for any C 2 function Q that satisfies the conditions 

dQ dQ >Q d 2 Q d 2 Q 
dK' dL ~ ' dK 2, dL 2 < 

If you consider what these relations mean qualitatively about the behavior of the 
output function with respect to increases in capital and labor, you will see that 
they are entirely reasonable assumptions. 4 ♦ 



4.4 Exercises 



1. Find the line that best fits the following data: (0, 2), 
(1,3), (2,5), (3,3), (4,2), (5,7), (6,7). 

2. Show that if you have only two data points (xi , y\ ) and 
(xi, yi), then the best fit line given by the method of 
least squares is, in fact, the line through (x\, yi) and 
(xz, yi)- 

3. Suppose that you are given « pairs of data (xi, y\), 
(X2, yz), ■ ■ ■ , (x„, y n ) and you seek to fit a function of 
the form y = a/x + b to these data. 

(a) Use the method of least squares as outlined in this 
section to construct a function D(a, b) that gives 
the sum of the squares of the distances between 
observed and predicted j-values of the data. 



(b) Show that the "best fit" curve of the form y = 
a/x + b should have 

n^yi/xt -(E !/*)(£*■) 

2 

»E i/*?-(£i/*) 

and 

b _ (E i/-*, 2 ) (E yj) - (E im) (E y.M-) 
»EV*?-(Ei/*) 2 

(All sums are from i = 1 to «.) 



4 For more about the history and derivation of the Cobb— Douglas function, consult R. Geitz, "The Cobb- 
Douglas production function," UMAP Module No. 509, Birkhauser, 1981. 



304 Chapter 4 | Maxima and Minima in Several Variables 



4. Find the curve of the form y = a/x + b that best fits 
the following data: (1 , 0), (2, - 1), (|, 1), and (3, - 1) . 
(See Exercise 3.) 

5. Suppose that you have n pairs of data (jci , ji), 
(xi, yi), ■ ■ ■ , (x n , y n ) and you desire to fit a quadratic 
function of the form y = ax 2 + bx + c to the data. 
Show that the "best fit" parabola must have coefficients 
a, b, and c satisfying 

(E xty + (E*, 3 ) & + (E-*, 2 ) c = Z*fyi 
(E -v, 3 )« + (E 4)b + (E *)c = E x m . 
(E^ 2 )« + (E^)^ + «c = E.v/ 

(All sums are from i = 1 to n.) 

6. (Note: This exercise will be facilitated by the use of 
a spreadsheet or computer algebra system.) Egbert 
recorded the number of hours he slept the night before 
a major exam versus the score he earned, as shown in 
the table below. 

(a) Find the line that best fits these data. 

(b) Find the parabola y = ax 2 + bx + c that best fits 
these data. (See Exercise 5.) 

(c) Last night Egbert slept 6.8 hr. What do your an- 
swers in parts (a) and (b) predict for his score on 
the calculus final he takes today? 



Hours of sleep 


Test score 


8 


85 


8.5 


72 


9 


95 


7 


68 


4 


52 


8.5 


75 


7.5 


90 


6 


65 



7. Let F = {-2x - 2y - l)i + (-2x - 6y - 2)j. 

(a) Show that F is conservative and has potential 
function 

V(x, y) = x 2 + 2xy + 3y 2 +x + 2y 

(i.e.,F = -VV). 

(b) What are the equilibrium points of F? The stable 
equilibria? 

8. Suppose a particle moves in a vector field F in R 2 with 
physical potential 

V(x, y) = 2x 2 - Sxy - y 2 + I2x - 8y + 12. 

Find all equilibrium points of F and indicate which, if 
any, are stable equilibria. 



9. Let a particle move in the vector field F in R 3 whose 
physical potential is given by 

V(x, y, z) = 3x 2 + 2xy + z 2 - 2yz + 3x + 5y - 10. 

Determine the equilibria of F and identify those that 
are stable. 

10. Suppose that a particle of mass m is constrained to 
move on the ellipsoid 2x 2 + 3y 2 + z 2 = 1 subject to 
both a gravitational force F = —mgk., as well as to an 
additional potential V(x, y, z) = 2x. 

(a) Find any equilibrium points for this situation. 

(b) Are there any stable equilibria? 

1 1 . The Sukolux Vacuum Cleaner Company manufactures 
and sells three types of vacuum cleaners: the standard, 
executive, and deluxe models. The annual revenue in 
dollars as a function of the numbers x, y, and z (re- 
spectively) of standard, executive, and deluxe models 
sold is 

R{x, y, z) = xyz 2 - 25,000* - 25,000y - 25,000z. 

The manufacturing plant can produce 200,000 total 
units annually. Assuming that everything that is manu- 
factured is sold, how should production be distributed 
among the models so as to maximize the annual 
revenue? 

1 2. Some simple electronic devices are to be designed to 
include three digital component modules, types 1, 2, 
and 3, which are to be kept in inventory in respective 
amounts X\, x%, and xj,. Suppose that the relative im- 
portance of these components to the various devices is 
modeled by the utility function 

U (x\ , X%, X3) = X\X2 + 2x\X^ + X\X2Xt,. 

You are authorized to purchase $90 worth of these parts 
to make prototype devices. If type 1 costs $1 per com- 
ponent, type 2 $4 per component, and type 3 $2 per 
component, how should you place your order? 

13. A farmer has determined that her cornfield will yield 
corn (in bushels) according to the formula 

B(x,y) = 4x 2 + y 2 + 600, 

where x denotes the amount of water (measured in 
hundreds of gallons) used to irrigate the field and y 
the number of pounds of fertilizer applied to the field. 
The fertilizer costs $10 per pound and water costs $15 
per hundred gallons. If she can allot $500 to prepare 
her field through irrigation and fertilization, use a 
Lagrange multiplier to determine how much water and 
fertilizer she should purchase in order to maximize her 
yield. 

14. A textile manufacturer plans to produce a cashmere/ 
cotton fabric blend for use in making sweaters. The 
amount of fabric that can be produced is given by 

f(x,y) = 4xy-2x-8y + 3, 



True/False Exercises for Chapter 4 



where x denotes the number of pounds of raw cash- 
mere used is and y is the number of pounds of raw 
cotton. Cotton costs $2 per pound and cashmere costs 
$8 per pound. 

(a) If the manufacturer can spend $ 1 000 on raw mate- 
rials, use a Lagrange multiplier to advise him how 
he should adjust the ratio of materials in order to 
produce the most cloth. 

(b) Now suppose that the manufacturer has a budget 
of B dollars. What should the ratio of cotton to 
cashmere be (in terms of 5)? What is the limiting 
value of this ratio as B increases? 

15. The CEO of the Wild Widget Company has decided 
to invest $360,000 in his Michigan factory. His eco- 
nomic analysts have noted that the output of this 
factory is modeled by the function Q(K, L) = 
60 K 1 ^ L 2 / 3 , where K represents the amount (in thou- 



sands of dollars) spent on capital equipment and L 
represents the amount (also in thousands of dollars) 
spent on labor. 

(a) How should the CEO allocate the $360,000 
between labor and equipment? 

(b) Check that BQ/dK = dQ/dL at the optimal 
values for K and L. 

16. Let Q{K,L) be a production function for a com- 
pany where K and L represent the respective amounts 
spent on capital equipment and labor. Let p denote the 
price of capital equipment per unit and w the cost of 
labor per unit. Show that, subject to a fixed produc- 
tion Q(K, L) = c, the total cost M of production is 
minimized when K and L are such that 

j_3G _ j_ae 

p dK w dL 



True/False Exercises for Chapter 4 



1 . If / is a function of class C 2 and p% denotes the second- 
order Taylor polynomial of / at a, then /(x) ~ P2W 
when x ~ a. 

2. The increment A/ of a function f(x, y) measures the 
change in the z-coordinate of the tangent plane to the 
graph of /. 

3. The differential df of a function f(x, y) measures the 
change in the z-coordinate of the tangent plane to the 
graph of /. 

4. The second-order Taylor polynomial of f(x, y, z) = 
x 2 + 3xz + y 2 at (1, —1, 2) is pi(x, y, z) = x 2 + 3xz 

+ y 2 - 

5. The second-order Taylor polynomial of f(x,y) = 
x 3 + 2xy + y at (0, 0) is pi(x, y) = 2xy + y. 

6. The second-order Taylor polynomial of f(x,y) = 
x 3 + 2xy + y at (1, —1) is p 2 (x, y) = 2xy + y. 

7. Near the point (1,3,5), the function f(x,y,z) = 
3x 4 + 2y 3 + z 2 is most sensitive to changes in z. 

8. The Hessian matrix Hf(x\ , . . . , x n ) of / has the prop- 
erty that Hf( Xl ,...,x n ) T = Hf(x x x„), 

9. If V/(ai , . . . , a n ) = 0, then / has a local extremum 
at a = (ai, . . . , a„). 

10. If / is differentiable and has a local extremum at 
a = (ai, . . . , a n ), then V/(a) = 0. 

11. The set {(x, y, z) \ 4 < x 2 + y 2 + z 2 < 9} is compact. 

12. The set {(x, y) \ 2x — 3y = 1} is compact. 



13. Any continuous function f(x, y) must attain a global 
maximum on the disk {(x, y) \ x 2 + y 2 < 1}. 

14. Any continuous function f(x,y,z) must attain a 
global maximum on the ball {(x, y, z) | (x — l) 2 + 
(y+l) 2 + z 2 <4). 

15. If f(x, y) is of class C 2 , has a critical point at (a, b), 
and f xx (a, b)f yy (a, b) — f xy (a, b) 2 < 0, then / has a 
saddle point at (a, b). 

16. If det ///(a) = 0, then / has a saddle point at a. 

17. The function f(x, y, z) = x i y 2 z — x 2 (y + z) has a 
saddle point at (1 , — 1 , 2). 

18. The function f(x, y, z) = x 2 + y 2 + z 2 — yz has a lo- 
cal maximum at (0, 0, 0). 

19. The function f(x, y, z) = xy 3 — x 2 z + z has a degen- 
erate critical point at (— 1 , 0, 0). 

20. The function F(x x ,...,.«„) = 2{x x - l) 2 - 3(x 2 - 2) 2 

_l_ 1_ (— iy i+1 ( n + l)(x n — rif has a critical point at 

(l,2,...,n). 

21 . The function F(xi, x„) = 2(xi- l) 2 - 3(x 2 - 2) 2 
+ ■ ■ ■ + (— l)" +1 (rc + l)(x n — n) 2 has a minimum at 
(l,2,...,n). 

22. All local extrema of a function of more than one vari- 
able occur where all partial derivatives simultaneously 
vanish. 

23. All points a = («i, . . . , ai) where the function 
f(x\, ... ,x„) has an extremum subject to the con- 
straint that g{x\, ... ,x n ) — c, are solutions to the 



Chapter 4 ! Maxima and Minima in Several Variables 



system of equations 

9/ 
3.vi 



9xi 



X - 



24. Any solution (Ai 
equations 

9*1 



9x„ 9x„ 
g(Xi, ...,x„) = c 

. . . , Xk,x\, . . . , x„) to the system of 



dx\ dx\ 



= Ai + • • • + Ajt 

dx n dx„ dx„ 

gi(xu ■■, X„) = C\ 



gl(xi, ...,X n ) = C k 

yields a point (x\, . . . , x n ) that is an extreme value 
of / subject to the simultaneous constraints g\ = 

a, ...,gk = c k . 

25. To find the critical points of the function f(x, y, z, w) 
subject to the simultaneous constraints g{x, y, z, w) = 



c, h(x, y , z, w) = d, k(x, y,z,w) = e using the tech- 
nique of Lagrange multipliers, one will have to solve 
a system of four equations in four unknowns. 

26. Suppose that f(x, y, z) and g(x, y, z) are of class C 1 
and that (xo , yo , zo) is a point where / achieves a maxi- 
mum value subject to the constraint that g(x , y , z) = c 
and that Vg(xo, yo, Zo) is nonzero. Then the level set 
of / that contains (xo, yo, zo) must be tangent to the 
level set S = {(x, y, z) \ g{x, y, z) = c}. 

27. The critical points of f(x, y, z) = xy + 2xz + 2yz 

subject to the constraint that xyz = 4 are the same 

as the critical points of the function F(x, y) = 

8 8 
xv + - + -. 

x y 

28. Given data points (3, 1), (4, 10), (5, 8), (6, 12), to find 
the best fit line by regression, we find the minimum 
value of the function D(m, b) = (3m + b — l) 2 + 
(4m + b - 10) 2 + (5m + b - 8) 2 + (6m + b - 12) 2 . 

29. All equilibrium points of a gradient vector field 
are minimum points of the vector field's potential 
function. 

30. Given an output function for a company, the marginal 
change in output per dollar investment in capital is the 
same as the marginal change in the output per dollar 
investment in labor. 



Miscellaneous Exercises for Chapter 4 



1. Let V = jtr 2 h, where r ~ ro and h ~ ho. What re- 
lationship must hold between ro and ho for V to be 
equally sensitive to small changes in r and hi 

2. (a) Find the unique critical point of the function 

f(x u x 2 , ...,x n ) = c -*?-*l— -< 

(b) Use the Hessian criterion to determine the nature 
of this critical point. 

3. The Java Joint Gourmet Coffee House sells top-of- 
the-line Arabian Mocha and Hawaiian Kona beans. 
If Mocha beans are priced at x dollars per pound 
and Kona beans at y dollars per pound, then mar- 
ket research has shown that each week approximately 
80 — lOOx + 40y pounds of Mocha beans will be sold 
and 20 + 60x — 35y pounds of Kona beans will be 
sold. The wholesale cost to the Java Joint owners is 
$2 per pound for Mocha beans and $4 per pound for 
Kona beans. How should the owners price the coffee 
beans in order to maximize their profits? 

4. The Crispy Crunchy Cereal Company produces three 
brands, X, Y, and Z, of breakfast cereal. Each month, 
x, y, and z (respectively) 1000-box cases of brands X, 



Y, and Z are sold at a selling price (per box) of each 
cereal given as follows: 



Brand 


No. cases sold 


Selling price per box 


X 


X 


4.00 - 0.02x 


Y 


y 


4.50-0.05y 


Z 


z 


5.00-O.lOz 



(a) What is the total revenue R if x cases of brand X, y 
cases of brand Y, and z cases of brand Z are sold? 

(b) Suppose that during the month of November, brand 
X sells for $3.88 per box, brand Y for $4.25, and 
brand Z for $4.60. If the price of each brand is in- 
creased by $0. 10, what effect will this have on the 
total revenue? 

(c) What selling prices maximize the total revenue? 

5. Find the maximum and minimum values of the 
function 

f(x, y, z) = x — V3 y 

on the sphere x 2 + y 2 + z 2 = 4 in two ways: 
(a) by using a Lagrange multiplier; 



Miscellaneous Exercises for Chapter 4 



the rectangle without overlapping, except along their 
edges. (See Figure 4.42.) 



(b) by substituting spherical coordinates (thereby de- 
scribing the point {x,y, z) on the sphere as x = 
2 sin (p cos 9, y = 2 sirup sin#, z = 2cos<p) and 
then finding the ordinary (i.e., unconstrained) ex- 
trema of f(x(cp, 9), y(<p, 9), z((p, 9)). 

6. Suppose that the temperature in a space is given by the 
function 

T(x, y,z) = 200xyz 2 . 

Find the hottest point(s) on the unit sphere in two ways: 

(a) by using Lagrange multipliers; 

(b) by letting x = sirup cos 6*, y = sirup sin 9, z = 
cos <p and maximizing T as a function of the two 
independent variables <p and 9 . (Note: It will help if 
you use appropriate trigonometric identities where 
possible.) 

7. Consider the function f(x, y) = (y — 2x 2 )(y — x z ). 

(a) Show that / has a single critical point at the origin. 

(b) Show that this critical point is degenerate. Hence, 
it will require means other than the Hessian crite- 
rion to determine the nature of the critical point as 
a local extremum. 

(c) Show that, when restricted to any line that passes 
through the origin, / has a minimum at (0, 0). 
(That is, consider the function F(x) = f(x, mx), 
where m is a constant and the function G(y) = 

f(0, y).) 

(d) However, show that, when restricted to the 
parabola y = |jc 2 , the function / has a global 
maximum at (0, 0). Thus, the origin must be a sad- 
dle point. 

(e) Use a computer to graph the surface z = f(x, y). 

8. (a) Find all critical points of f(x, y) = xy that satisfy 

x 2 + y 2 = 1. 

(b) Draw a collection of level curves of / and, on the 
same set of axes, the constraint curve x 2 + y 2 = 1 , 
and the critical points you found in part (a). 

(c) Use the plot you obtained in part (b) and a geomet- 
ric argument to determine the nature of the critical 
points found in part (a). 

9. (a) Find all critical points of f(x,y,z) = xy that 

satisfy x 2 + y 2 + z 2 = 1 . 

(b) Give a rough sketch of a collection of level surfaces 
of / and, on the same set of axes, the constraint 
surface x 2 + y 2 + z 2 = 1 , and the critical points 
you found in part (a). 

(c) Use part (b) and a geometric argument to determine 
the nature of the critical points found in part (a). 

10. Find the area A of the largest rectangle so that two 
squares of total area 1 can be placed snugly inside 



Figure 4.42 Figure for Exercise 10. 

1 1 . Find the minimum value of 

f{x\ ,x 2 , ... , x„) = x\ + x\ H h x 2 

subject to the constraint that a\X\ + a 2 x 2 + • • • + 
a„x n = 1, assuming that a\ + a\ + • • • + a 2 > 0. 

12. Find the maximum value of 

f(x u x 2 , ...,x n ) = (aiXi + a 2 x 2 H h a n x n ) 2 , 

subject to x\ + x\ + • • • + x 2 = I. Assume that not all 
of the a, 's are zero. 

1 3. Find the dimensions of the largest rectangular box that 
can be inscribed in the ellipsoid x 2 + 2y 2 + 4z 2 = 12. 
Assume that the faces of the box are parallel to the 
coordinate planes. 

14. Your company must design a storage tank for Super 
Suds liquid laundry detergent. The customer's specifi- 
cations call for a cylindrical tank with hemispherical 
ends (see Figure 4.43), and the tank is to hold 8000 gal 
of detergent. Suppose that it costs twice as much (per 
square foot of sheet metal used) to machine the hemi- 
spherical ends of the tank as it does to make the cylin- 
drical part. What radius and height do you recommend 
for the cylindrical portion so as to minimize the total 
cost of manufacturing the tank? 




Figure 4.43 The storage tank 
of Exercise 14. 

15. Find the minimum distance from the origin to the 
surface x 2 — (y — z) 2 = 1 . 

1 6. Determine the dimensions of the largest cone that can 
be inscribed in a sphere of radius a. 

17. Find the dimensions of the largest rectangular box 
(whose faces are parallel to the coordinate planes) that 



Chapter 4 ! Maxima and Minima in Several Variables 



can be inscribed in the tetrahedron having three faces in 
the coordinate planes and fourth face in the plane with 
equation bcx + acy + abz = abc, where a, b, and c 
are positive constants. (See Figure 4.44.) 




Figure 4.44 Figure for Exercise 17. 

1 8. You seek to mail a poster to your friend as a gift. You 
roll up the poster and put it in a cylindrical tube of di- 
ameter a' and length y. The postal regulations demand 
that the sum of the length of the tube plus its girth (i.e., 
the circumference of the tube) be at most 108 in. 

(a) Use the method of Lagrange multipliers to find the 
dimensions of the largest- volume tube that you can 
mail. 

(b) Use techniques from single-variable calculus to 
solve this problem in another way. 

1 9. Find the distance between the line y = 2x + 2 and the 
parabola x = y 2 by minimizing the distance between 
a point (x\ , yi) on the line and a point (X2 , yi) on the 
parabola. Draw a sketch indicating that you have found 
the minimum value. 

20. A ray of light travels at a constant speed in a uniform 
medium, but in different media (such as air and water) 
light travels at different speeds. For example, if a ray of 
light passes from air to water, it is bent (or refracted) 
as shown in Figure 4.45. Suppose the speed of light 

A 



\^ Medium 


1 


a i V 




i V 1 ! 




h 




i 


i 


Medium 2 10 £ 


i 




B 



in medium 1 is v\ and in medium 2 is 1)2 ■ Then, by 
Fermat's principle of least time, the light will strike the 
boundary between medium 1 and medium 2 at a point 
P so that the total time the light travels is minimized. 

(a) Determine the total time the light travels in going 
from point A to point B via point P as shown in 
Figure 4.45. 

(b) Use the method of Lagrange multipliers to estab- 
lish Snell's law of refraction: that the total travel 
time is minimized when 



sin 6*i 
sin 6*2 



Vl 
!'2 ' 



(Hint: The horizontal and vertical separations of 
A and B are constant.) 

21. Use Lagrange multipliers to establish the formula 
\ax 0 + by 0 — d\ 



D 



V« 2 + b 2 



for the distance D from the point (xq, yo) to the line 
ax + by = d. 

22. Use Lagrange multipliers to establish the formula 

\ax 0 + by 0 + czo - d\ 



D 



si a 2 + b 2 + c 2 



for the distance D from the point (xq, yo, zo) to the 
plane ax + by + cz = d. 



23. (a) Show that the maximum value of f(x,y,z) 

2 2' 

x A y A z 



■ 2 subject to the constraint that x 2 + y 2 + 



2 2 
Z = a is 



27 V 3 

(b) Use part (a) to show that, for all x, y, and z, 



(*Vz 2 ) 1/3 



x 2 + y 2 + z 2 



(c) Show that, for any positive numbers x\ , x% , . . . , x n , 

i ^ x l +x 2 + ---+x n 

n 

The quantity on the right of the inequality is the 
arithmetic mean of the numbers x\, X2, ■ ■ . , x n , 
and the quantity on the left is called the geomet- 
ric mean. The inequality itself is, appropriately, 
called the arithmetic-geometric inequality. 

(d) Under what conditions will equality hold in the 
arithmetic-geometric inequality? 

In Exercises 24-27 you will explore how some ideas from ma- 
trix algebra and the technique of Lagrange multipliers come 
together to treat the problem of finding the points on the unit 
hypersphere 



Figure 4.45 Snell's law of refraction. 



g(xu ■■■,x„) 



+ x 2 + - 



l 



4.6 | Miscellaneous Exercises for Chapter 4 



that give extreme values of the quadratic form 
f(*L 



where the Ojj s are constants. 



24. (a) Use a Lagrange multiplier X to set up a system of 
n + 1 equations in w + 1 unknowns xi , . . . , x n , X 
whose solutions provide the appropriate con- 
strained critical points. 

(b) Recall that formula (2) in §4.2 shows that the 
quadratic form / may be written in terms of ma- 
trices as 



f(xi 



,x„) 



(1) 



where the vector x is written as the « x 1 matrix 

X\ 

and A is the n x n matrix whose ij th entry 

is atj . Moreover, as noted in the discussion in §4.2, 
the matrix A may be taken to be symmetric (i.e., so 
that A T = A), and we will therefore assume that 
A is symmetric. Show that the gradient equation 
V/ = XV g is equivalent to the matrix equation 



Ax = Xx. 



(2) 



Since the point (xi, . . . , x n ) satisfies the constraint 
*]+•••+ x% = 1, the vector x is nonzero. If you 
have studied some linear algebra, you will recog- 
nize that you have shown that a constrained criti- 
cal point (xi, . . . , x„) for this problem corresponds 
precisely to an eigenvector of the matrix A asso- 
ciated with the eigenvalue X. 



(c) Now suppose that x : 



x\ 



is one of the eigen- 



vectors of the symmetric matrix A, with associated 
eigenvalue X. Use equations (1) and (2) to show, if 
x is a unit vector, that 



/(*! 



■ , X n ) = X. 



Hence, the (absolute) minimum value that / 
attains on the unit hypersphere must be the small- 
est eigenvalue of A and the (absolute) maximum 
value must be the largest eigenvalue. 

25. Let n = 2 in the situation of Exercise 24, so that we are 
considering the problem of finding points on the circle 



x 2 + y 2 = 1 that give extreme values of the function 

f(x, y) = ax 2 + 2bxy + cy 2 



[* y] 



(a) Find the eigenvalues of A 



a b 
b c 



by identify- 



ing the constrained critical points of the optimiza- 
tion problem described above. 

(b) Now use some algebra to show that the eigenval- 
ues you found in part (a) must be real. It is a fact 
(that you need not demonstrate here) that any n x n 
symmetric matrix always has real eigenvalues. 

26. In Exercise 25 you noted that the eigenvalues X\, X2 
that you obtained are both real. 

(a) Under what conditions does X\ = X2I 

(b) Suppose that X\ and X2 are both positive. Explain 
why / must be positive on all points of the unit 
circle. 

(c) Suppose that Xi and X2 are both negative. Explain 
why / must be negative on all points of the unit 
circle. 

27. Let / be a general quadratic form in n variables de- 
termined by an n x n symmetric matrix A, that is, 

f{x\, x„) = a u x i x j = x r ^ x - 

(a) Show, for any real number k, that f(kx\,..., 
kx„) = k 2 f(x\, . . . , x„). (This means that a 
quadratic form is a homogeneous polynomial of 
degree 2 — see Exercises 37-44 of the Miscella- 
neous Exercises for Chapter 2 for more about ho- 
mogeneous functions.) 

(b) Use part (a) to show that if / has a positive 
minimum on the unit hypersphere, then / must 
be positive for all nonzero x e R" and that if / 
has a negative maximum on the unit hypersphere, 
then / must be negative for all nonzero x € R" . 
(Hint: For x / 0, let u = x/||x||, so that x = ku, 
where k = ||x||.) 

(c) Recall from §4.2 that a quadratic form / is said 
to be positive definite if /(x) > 0 for all nonzero 
x e R" and negative definite if /(x) < 0 for all 
nonzero x e R" . Use part (b) and Exercise 24 to 
show that the quadratic form / is positive definite 
if and only if all eigenvalues of A are positive, and 
negative definite if and only if all eigenvalues of 
A are negative. (Note: As remarked in part (b) of 
Exercise 25, all the eigenvalues of A will be real.) 




Multiple Integration 



5.1 Introduction: Areas and 
Volumes 

5.2 Double Integrals 

5.3 Changing the Order of 
Integration 

5.4 Triple Integrals 

5.5 Change of Variables 

5.6 Applications of Integration 

5.7 Numerical Approximations 
of Multiple Integrals 
(optional) 

True/False Exercises for 
Chapter 5 

Miscellaneous Exercises for 
Chapter 5 

y 



Figure 5.1 The graph of 

y = f(x). 
y 



Figure 5.2 The shaded region has 
area /* f(x)dx. 



5.1 Introduction: Areas and Volumes 

Our purpose in this chapter is to find ways to generalize the notion of the definite 
integral of a function of a single variable to the cases of functions of two or 
three variables. We also explore how these multiple integrals may be used to 
meaningfully represent various physical quantities. 

Let / be a continuous function of one variable defined on the closed interval 
[a, b] and suppose that / has only nonnegative values. Then the graph of / looks 
like Figure 5.1. That / is continuous is reflected in the fact that the graph consists 
of an unbroken curve. That / is nonnegative-valued means that this curve does not 
dip below the x -axis. We know from one- variable calculus that the definite integral 
f b f(x) dx exists and gives the area under the curve, as shown in Figure 5.2. 

Now suppose that / is a continuous, nonnegative-valued function of two 
variables defined on the closed rectangle 

R = [( x , y) e R 2 | a < x < b, c < y < d] 

in R 2 . Then the graph of / over R looks like an unbroken surface that never dips 
below the xy-plane, as shown in Figure 5.3. In analogy with the single-variable 
case, there should be some sort of integral that represents the volume under the 
part of the graph that lies over R. (See Figure 5.4.) We can find such an integral 
by using Cavalieri's principle, which is nothing more than a fancy term for the 
method of slicing. Suppose we slice by the vertical plane x = xo, where xo is a 
constant between a and b. Let A(xo) denote the cross-sectional area of such a 
slice. Then, roughly, one can think of the quantity A(xo) dx as giving the volume 
of an "infinitely thin" slab of thickness dx and cross-sectional area A(xo). (See 
Figure 5.5.) Hence, the definite integral 



V 



-f 

J Li 



A(x) dx 



gives a "sum" of the volumes of such slabs and can be considered to provide a 
reasonable definition of the total volume of the solid. 

But what about the value of A(xo)? Note that A(xo) is nothing more than the 
area under the curve z = f(xo, y), obtained by slicing the surface z = f(x, y) 
with the plane x = xq. Therefore, 



A(x 0 ) 



f(x 0 , y)dy 



5.1 | Introduction: Areas and Volumes 311 




Figure 5.3 The graph of Figure 5.4 The region under the Figure 5.5 A slab of "volume" 

z = f(x, v). portion of the graph of / lying dV = A(x 0 ) dx. 

over R has volume that is given 

by an integral. 



Plane z = c 



(a, 0, 0) 




(0,6,0) 

y 

(a, b, 0) 



Figure 5.6 Calculating the 
volume of the box of Example 1 . 




Figure 5.7 The graph of 
z = 4 - x 2 
Example 2. 



4 - x 2 - y 2 of 



(remember xq is a constant), and so we find that 



o pupa 

= / A(x)(ix = / / /(x, y)dy 

J a J a lJ c 



dx. 



(i) 



The right-hand side of formula (1) is called an iterated integral. To calculate 
it, first find an "antiderivative" of f(x, y) with respect to y (by treating i as a 
constant), evaluate at the integration limits y = c and y = d, and then repeat the 
process with respect to x. 

EXAMPLE 1 Let's make sure that the iterated integral defined in formula (1) 
gives the correct answer in a case we know well, namely, the case of a box. We'll 
picture the box as in Figure 5.6. That is, the box is bounded on top and bottom by 
the planes z = c (where c > 0) and z = 0, on left and right by the planes y = 0 
and y = b (where b > 0), and on back and front by the planes x = 0 and x = a 
(a > 0). Hence, the volume of the box may be found by computing the volume 
under the graph of z = c over the rectangle 

R = {(x,y) | 0 < x < a,0 < y < b}. 

Using formula (1), we obtain 

V = y j cdydx = j (^cy\- y ~^ dx = j cbdx = cbx\]z C Q = cba. 

This result checks with what we already know the volume to be, as it should. ♦ 

EXAMPLE 2 We calculate the volume under the graph of z = 4 — x 2 — y 2 

(Figure 5.7) over the square 

R = {(x, y) | -1 < x < 1, -1 < y < 1}. 

Using formula (1) once again, we calculate the volume by first integrating with 
respect to y (i.e., by treating x as a constant in the inside integral) and then by 



Chapter 5 | Multiple Integration 



y = y 0 plane 




gure 5.8 

first. 



Y/y 

Slicing by y = yo 



integrating w 



■/. 



th respect to x. The details are as follows: 
j (4-x 2 - y 2 )dydx = j 



dx 



y=-l 



A-x z 

8 - 2x 2 
22 2 : 

X X 

3 3 



-4 + x 2 + 



dx 



dx 



-l 



22 
T 



22 2 
T + 3 



40 

y 



In our development of formula (1 ), we could just as well have begun by slicing 
the solid with the plane y = yo (instead of with the plane x = xq), as shown in 
Figure 5.8. Then, in place of formula (1), the formula that results is 



V 



rd rb 

= I I f( x , y )dxdy. 

J c J a 



(2) 



Since the iterated integrals in formulas ( 1 ) and (2) both represent the volume of the 
same geometric object, we can summarize the preceding discussion as follows. 



PROPOSITION 1 .1 Let R be the rectangle {(x, y) \ a < x < b, c < y < d) and 
let / be continuous and nonnegative on R. Then the volume V under the graph 
of /' over R is 

pb pel pel pb 

/ / f(x,y)dydx= / f(x,y)dxdy. 

J a J c J c J a 



EXAMPLE 3 We find the volume under the graph of z = cosx siny over the 
rectangle 



R 



f 7T TV 1 

= \(x,y) \ 0<x <-, 0<y <-\ 



(See Figure 5.9.) From formula (1), we calculate that the volume is 



V 



fJt/2 fx/2 

= / cos x sin y dy dx = / 

Jo Jo Jo 



*' 2 ( sj2_ 
2 



(— cosx cos j)|y = o /4 dx 



2-V2 r 12 



— 2~ — J 



2- s/2 



sinx 



7T/2 



2 - V2 2 - s/2 

—^(1-0) = —^-. 



5.1 | Exercises 313 




Figure 5.9 The surface z = cosx siny of 
Example 3. 



If we use formula (2) instead of formula (1), we obtain 

tjt/4 fji/2 r ir/4 



Jo Jo 



I cos x sin y dx dy = / (sinx sin y)\ x x= ^ 2 dy 
Jo Jo 

*n/4 



= f 

Jo 



,t/4 



(sin y — 0) dy = — cos y | 0 

4l 2-yfl 

= ~- (-1) = — • 

2 V ' 2 

That this result agrees with our first calculation is no surprise given 
Proposition 1.1. ♦ 



5.1 Exercises 



Evaluate the iterated integrals given in Exercises 1-6. 



2. J J y sin x dy dx 

3. / / xe y dy dx 

J -2 Jo 

rn/2 ri 

4. / / e x cos y dx dy 
Jo Jo 

5. J j (e x+y + x 2 + In y) dx dy 

6. f f — ^— dx dy 

Jl Ji xy 



7. Find the volume of the region that lies under the graph 
of the paraboloid z = x 2 + y 2 +2 and over the rect- 
angle R = {(x, y) | — 1 < x < 2, 0 < y < 2} in two 
ways: 

(a) by using Cavalieri's principle to write the volume 
as an iterated integral that results from slicing the 
region by parallel planes of the form x = constant; 

(b) by using Cavalieri's principle to write the volume 
as an iterated integral that results from slicing the 
region by parallel planes of the form y = constant. 

8. Find the volume of the region bounded on top by the 
plane z = x + 3y + 1, on the bottom by the xy-plane, 
and on the sides by the planes x = 0, x = 3, y = 1, 

y = 2. 

9. Find the volume of the region bounded by the graph 
of f{x, y) = 2x 2 + y 4 sinjrx, the xy-plane, and the 
planes x = 0, x = 1, y = — 1, y = 2. 



314 Chapter 5 | Multiple Integration 



In Exercises 10—15, calculate the given iterated integrals and 
indicate of what regions in R 3 they may be considered to 
represent the volumes. 

10. j j Idxdy 

11. J j (16-x 2 -y 2 )dydx 

/jr/2 />jt 
I sin x cos y dx dy 
-jr/2 JO 



13. j j (4-x 2 )dxdy 



14. J J \x\ sin jry dy dx 

15. J J (5-\y\)dxdy 

16. Suppose that / is a nonnegative- valued, continu- 
ous function defined on R = {(x, y) | a < x < b, c < 
y < d}. If fix, y) < M for some positive number M, 
explain why the volume V under the graph of / over 
R is at most M(b — a)(d — c). 



5.2 Double Integrals 

In the previous section we saw how to calculate volumes of certain solids us- 
ing iterated integrals. The ideas were mostly straightforward, but the situation 
we addressed was rather special: We only solved the problem of computing the 
volume of a solid defined as the region lying under the graph of a continuous, 
nonnegative-valued function f(x, y) and above a rectangle in the xv-plane. It is 
not immediately apparent how we might compute the volume of a more general 
solid based on this work. 

Thus, in this section we define a more general notion of an integral of a 
function of two variables that will allow us to describe 

1. integrals of arbitrary functions (i.e., functions that are not necessarily non- 
negative or continuous) and 

2. integrals over arbitrary regions in the plane (i.e., rather than integrals over 
rectangles only). 

We focus first on case 1 . To do this, we start fresh with some careful definitions 
and notation. The ideas involved in Definitions 2.1-2.3 below are different from 
those in the previous section. However, we will see that there is a key connection 
(called Fubini's theorem) between the notion of an iterated integral discussed 
in §5.1 and that of a double integral, which will be described in Definition 2.3. 

The Integral over a Rectangle 

We also denote a (closed) rectangle 

R = {{x, y) e R 2 | a < x < b, c < y < d} 

by [a, b] x [c, d]. This notation is intended to be analogous to the notation for a 
closed interval. 



DEFINITION 2.1 Given a closed rectangle R = [a, b] x [c, d], a partition 
of R of order n consists of two collections of partition points that break up 
R into a union of n 2 subrectangles. More specifically, for i, j = 0, . . . , n, we 
introduce the collections {*, } and {yj}, so that 

a = xq < x\ < ■ ■ ■ < x,_i < Xj < ■ ■ ■ < x„ = b, 



5.2 | Double Integrals 315 



and 



c = yo < yi < 



< 



yj-y < yj 



< y n = d. 



Let Ax,- = x,- — Xi-i (for i = 1, . . . , n) and Ay ; - = y,- — y;_i (for j = 
1, . . . , n). Note that Ax,- and Ay v - are just the width and height (respec- 
tively) of the z'y'th subrectangle (reading left to right and bottom to top) of the 
partition. 



An example of a partitioned rectangle is shown in Figure 5.10. We do not 
assume that the partition is regular (i.e., that all the subrectangles have the same 
dimensions). 



d = y„ 

yj-i 
c=y 0 



+ 



H h 



+ 



+ 



+ 



CI — Xq X-[ X2 ' ■ ■ Xj _ 1 Xj • • • X n — D 

Figure 5.10 A partition of the rectangle [a, b] x [c, d]. 



DEFINITION 2.2 Suppose that / is any function denned on R = [a, b] x 
[c, d] and partition R in some way. Let c, ; - be any point in the subrectangle 

Rij = [xi-\,Xi\ x [yj-uyj] (i,j = l,..., n). 

Then the quantity 

n 

where AAjj = Ax, Ay 7 is the area of Rij, is called a Riemann sum of / on 
R corresponding to the partition. 



The Riemann sum 

S = £/(cy)AA (7 

ij 

depends on the function /, the choice of partition, and the choice of the "test 
point" cy in each subrectangle Rij of the partition. The Riemann sum itself is 
just a weighted sum of areas AAjj of subrectangles of the original rectangle R, 
the weighting being given by the value /(cy). 

If / happens to be nonnegative on R, then, for i, j = 1 , . . . , n, the individual 
terms /(cy ) A Ay in S may be considered to be volumes of boxes having base area 



Chapter 5 | Multiple Integration 



z 




Figure 5.11 The volume under the graph Figure 5.12 The Riemann sum as a signed sum of 

of / is approximated by the Riemann sum. volumes of boxes. 



Ax, Ay j and height /(cy ). Therefore, S can be considered to be an approximation 
to the volume under the graph of / over R, as suggested by Figure 5.1 1. If / is 
not necessarily nonnegative, then the Riemann sum S is a signed sum of such 
volumes (because, with /(cy) < 0, the term /(Cij)AA,j is the negative of the 
volume of the appropriate box — see Figure 5.12). 



Al\ 









Figure 5.13 HA U A 2 ,A 3 
represent the values of the shaded 
areas, then 

f* f(x)dx = A l -A 2 + A 3 . 




R 

(in xy-plane) 

Figure 5.14 If V\, V 2 represent 
the volumes of the shaded regions, 
then ff R f(x, y)d A = V l -V 2 . 



DEFINITION 2.3 The double integral of / on R, denoted by ff R f dA 
(or by f f R f{x, y) dA or by / f R f(x, y) dx dy), is the limit of the Riemann 
sum S as the dimensions Ax; and Ayj of the subrectangles Rjj all approach 
zero, that is, 

f f fdA = lim V f(Cij)AxiAyj, 

provided of course, that this limit exists. When ff R f dA exists, we say that 
/ is integrable on R. 



The crucial idea to remember — indeed, the defining idea — is that the integral 
ff R f dA is a limit of Riemann sums S, for this concept is what is needed to 
properly apply double integrals to physical situations. 

From a geometric point of view, just as the single-variable definite integral 
fa f( x ) dx can be used to compute the "net area" under the graph of the curve 
v = f(x) (as in Figure 5. 13), the double integral ff R fdA can be used to compute 
the "net volume" under the graph of z = fix, y) (as in Figure 5.14). 

Another way to view the double integral ff R f dA is somewhat less geometric 
but is more in keeping with the notion of the integral as the limit of Riemann 
sums and provides a perspective that generalizes to triple integrals of functions of 
three variables. Instead of visualizing the graph of z = fix, y) as a surface and 
S = Yll j=\ f( c ij)AxiAyj as a (signed) sum of volumes of boxes related to the 
graph, consider S to be a weighted sum of areas and the integral ff R f dA the 
limiting value of such weighted sums as the dimensions of all the subrectangles 
approach zero. With this point of view, we do not depict the integrand / when we 
try to visualize the integral. In this way, the distinction between the roles of the 
integrand and the rectangle R over which we integrate can be made clearer. (See 
Figure 5.15.) 



5.2 | Double Integrals 317 



R 



- This subrectangle of 
area A^4 53 contributes 
/(c 53 ) AA 53 to S. 



Figure 5.15 S = /(Cy)AAy. 



EXAMPLE 1 Suppose that a 3 cm square metal plate is made, but some nonuni- 
formities exist due to the manufacturing process so that the mass density varies 
somewhat throughout the plate. If we knew the density function 8(x, y) at every 
point in the plate, then we could calculate the total mass of the plate as 

Total mass = j j S(x,y)dA, 

where D denotes the square region of the plate placed in an appropriate coordinate 
system. 

In the absence of an analytic expression for S , we nonetheless can approximate 
the double integral by means of a Riemann sum: We partition the square region 
of the plate, take density readings at a test point in each subregion, and combine 
to approximate the integral for the total mass. (Essentially what we are doing is 
assuming that the density is nearly constant on each subregion so that multiplying 
density and area will give the approximate mass of the subregion; adding these 
approximate masses then gives an approximation for the total mass.) For example, 
we might model the problem as in Figure 5.16, where the region of the plate is 



(0.6) 



(0.3) 



(0.2) 



(0.2) 



(0.1) 



(0.5) 



(0.3) 



(0.3) 



Figure 5.16 The region of Example 1. The 3x3 
square is partitioned into nine subregions. The density 
values at test points in each subregion are shown. 



Chapter 5 | Multiple Integration 




Figure 5.17 The graph 
of z = x of Example 2. 



HI 



Figure 5.1 8 The two 

subrectangles R^j and Ri 2 j are 
symmetrically placed with respect 
to the y-axis. The corresponding 
test points C;, / and c> 2 j are chosen 
so that they have the same 
y-coordinates and opposite 
x-coordinates. 



partitioned into nine square subregions. Then we have 
Total mass = j j S(x, y)dA «a ^ 5(c, 7 )AA y 

= (0.2)1 + (0.3)1 + (0.6)1 + (0.1)1 + (0.2)1 + (1)1 + (0.3)1 

+ (0.3)1 +(0.5)1 = 3.5. ^ 

EXAMPLE 2 We determine the value of ff R xdA, where R = [-2,2] x 
[—1, 3]. Here the integrand f(x, y) = x and, if we graph z = f(x, y) over R, 
we see that we have a portion of a plane, as shown in Figure 5.17. Note that the 
portion of the plane is positioned so that exactly half of it lies above the xy-plane 
and half below. Thus, if we regard ff R x dA as the net volume under the graph of 
z = x, then we conclude that ff R x dA (if it exists) must be zero. 

On the other hand, we need not resort to visualization in three dimensions. 
Consider a Riemann sum corresponding to ff R x dA obtained by partitioning 
R = [—2, 2] x [—1,3] symmetrically with respect to the y-axis and by choosing 
the "test points" c, ; symmetrically also. (See Figure 5.18.) It follows that the value 
of 

S = f( c u) AA ij = J2 Xi i AAi J 

(where x, 7 denotes the x-coordinate of cy) must be zero since the terms of the 
sum cancel in pairs. Furthermore, we can arrange things so that, as we shrink 
the dimensions of the subrectangles to zero (as we must do to get at the integral 
itself), we preserve all the symmetry just described. Hence, the limit under these 
restrictions will be zero, and thus, the overall limit (where we do not impose such 
symmetry restrictions on the Riemann sum), if it exists at all, must be zero as 
well. ♦ 

Example 2 points out fundamental difficulties with Definition 2.3, namely, 
that we never did determine whether ff R f dA really exists. To do this, we would 
have to be able to calculate the limit of Riemann sums of / over all possible 
partitions of R by using all possible choices for the test points c,y, a practically 
impossible task. Fortunately, the following result (which we will not prove) pro- 
vides an easy criterion for integrability: 




Figure 5.1 9 The graph of a 
piecewise continuous function. 



THEOREM 2.4 If / is continuous on the closed rectangle R, then ff R f dA 
exists. 

In Example 2, f(x, y) = x is a continuous function and hence integrable 
by Theorem 2.4. The symmetry arguments used in the example then show that 
ff R xdA = 0, 

Continuous functions are not the only examples of integrable functions. In 
the case of a function of a single variable, piecewise continuous functions are also 
integrable. (Recall that a function f(x) is piecewise continuous on the closed 
interval [a, b] if / is bounded on [a, b] and has at most finitely many points of 
discontinuity on the interior of [a , b] . Its graph, therefore, consists of finitely many 
continuous "chunks" as shown in Figure 5.19.) For a function of two variables, 
there is the following result, which generalizes Theorem 2.4. 



5.2 | Double Integrals 319 



THEOREM 2.5 If / is bounded on R and if the set of discontinuities of / on R 
has zero area, then ff R f dA exists. 



To say that a set X has zero area as we do in Theorem 2.5, we mean that we 
can cover X with rectangles R\, R2, . . . , R ni . . . (i.e., so that X c [J^Lj ^«)> tne 
sum of whose areas can be made arbitrarily small. 

A function / satisfying the hypotheses of Theorem 2.5 has a graph that 
looks roughly like the one in Figure 5.20. Theorem 2.5 is the most general suffi- 
cient condition for integrability that we will consider. It is of particular use to us 
when we define the double integral of a function over an arbitrary region in the 
plane. 



a a 




X 

Discontinuities of/ 



Figure 5.20 The graph of an integrable 
function. 



Although Theorems 2.4 and 2.5 make it relatively straightforward to check 
that a given integral exists, they do little to help provide the numerical value of 
the integral. To mechanize the evaluation of double integrals, we will use the 
following result: 



THEOREM 2.6 (Fubini's theorem) Let / be bounded on R = [a, b] x [c, d] 
and assume that the set S of discontinuities of / on R has zero area. If every line 
parallel to the coordinate axes meets S in at most finitely many points, then 

p p pb pd pd pb 

// fdA= / / f(x,y)dydx= / / f(x,y)dxdy. 

J J R J a J c J c J a 



Fubini's theorem demonstrates that under certain assumptions the double in- 
tegral over a rectangle (i.e., the limit of Riemann sums) can be calculated by using 
iterated integrals and, moreover, that the order of integration for the iterated inte- 
gral does not matter. We remark that the independence of the order of integration 
depends strongly on the fact that the region of integration is rectangular; it will not 



320 Chapter 5 | Multiple Integration 



generalize to more arbitrary regions in such a simple way. (A proof of Theorem 
2.6 is given in the addendum to this section.) 

EXAMPLE 3 Werevisit ff R x dA in Example 2, where R = [-2, 2] x [-1, 3]. 
By Theorem 2.6, we know that Jf R xdA exists and by Fubini's theorem, we 
calculate 



j j xdA = j j x dy dx = j 



xy 



=-i 



dx 



-L 



x(3 - {-\))dx 



/: 



Ax dx = 2x 



2|2 

1-2 



8 = 0, 



which checks. Furthermore, we also have 



j j x dA = j j xdxdy = j 

J JR J-l J-2 J-] 



dy = f_ l (2-2)dy = 0. 

x=—2 



PROPOSITION 2.7 (Properties of the integral) Suppose that / and g are 
both integrable on the closed rectangle R. Then the following properties hold: 



1. / + g is also integrable on R and 



/ / (f + g)dA= f f fdA+ f f gdA. 
J Jr J J r J J r 



2. cf is also integrable on R, where c e R is any constant, and 



\L cfiA = c \l SdA ' 



3. If f(x, y) < g(x, y) for all (x,y)e R, then 



If f{x,y)dA< I j g(x,y)dA. 



4. | f\ is also integrable on R and 



f Ir f Jr 



< / / \f\dA. 



Properties 1 and 2 are called the linearity properties of the double integral. 
They can be proved by considering the appropriate Riemann sums and taking 
limits. For example, to prove property 1, note that the Riemann sum whose limit 



5.2 | Double Integrals 321 



is 



ff R (f + g)dA 



is 



Figure 5.21 

in the plane. 



A bounded region D 




Figure 5.22 A type 1 elementary 



region. 




) 


> 


y = d 










POO 








y = c 



Figure 5.23 A type 2 elementary 
region. 



t'J=i 



= E /( c v)AA !7 + E *(c/y)AAy 

U I 1,7 = 1 



Property 3 (known as monotonicity) and property 4 can also be proved using 
Riemann sums. For property 4, one needs to use the fact that 



E«* 

k=l 



<E |a *!- 

fc=l 



Double Integrals over General Regions in the Plane 

Our next step is to understand how to define the integral of a function over an 
arbitrary bounded region D in the plane. Ideally, we would like to give a precise 
definition of ff D fdA, where D is the amoeba-shaped blob shown in Figure 5.21 
and where / is bounded on D. In keeping with the definition of the integral over 
a rectangle, ff D f dA should be a limit of some type of Riemann sum and should 
represent the net volume under the graph of / over D. Unfortunately, the techni- 
calities involved in making such a direct approach work are prohibitive. Instead, we 
shall consider only certain special regions (rather than entirely arbitrary ones), and 
we shall assume that the integrand / is continuous over the region of integration 
(which will allow us to use what we already know about integrals over rectangles). 
Although this approach will not provide us with a completely general definition, 
it is sufficient for essentially all the practical situations we will encounter. 
To begin, we define the types of elementary regions we wish to consider. 



DEFINITION 2.8 We say that D is an elementary region in the plane if it 
can be described as a subset of R 2 of one of the following three types: 

Type 1 (see Figure 5.22): 

D = {(*, y) | y(x) <y< S(x), a<x <b}, 
where y and <5 are continuous on [a, b]. 
Type 2 (see Figure 5.23): 

D = {(x, y) | a(y) < x < 0(y), c<y<d}, 
where a and /3 are continuous on [c, d]. 
Type 3 D is of both type 1 and type 2. 



Thus, a type 1 elementary region D has a boundary (denoted 3D) consisting 
of straight segments (possibly single points) on the left and on the right and graphs 
of continuous functions of x on the top and on the bottom. A type 2 elementary 



322 Chapter 5 | Multiple Integration 



region has a boundary that is straight on the top and bottom and consists of graphs 
of continuous functions of y on the left and right. 

EXAMPLE 4 The unit disk, shown in Figure 5.24, is an example of a type 3 
elementary region. It is a type 1 region since 



D = Ux,y) | — v/l - x 2 



y < yi — . 

(See Figure 5.25.) It is also a type 2 region since 



D = Ux, y) | —/l - y 2 <x< 
(See Figure 5.26.) 



y 



-1 <x < lj . 

i<y<i}- 




Figure 5.24 The unit disk 
D = {(x, y) | a 2 + v 2 < 1} is a 
type 3 region. 






y = l 




/ 
















y = -l 



Figure 5.25 The unit disk D as a 
type 1 region. 



Figure 5.26 The unit disk D as a type 2 
region. 



Now we are ready to define ff D f dA, where D is an elementary region and 
/ is continuous on D. We construct a new function / ext , the extension of /, by 



f ex \x,y) = 



f(x,y) if (x,y)eD 
0 if(x,y)£D 



Note that, in general, / ext will not be continuous, but the discontinuities of / ext 
will all be contained in 3D, which has no area. Hence, by Theorem 2.5, / ext is 
integrable on any closed rectangle R that contains D. (See Figure 5.27.) 



Figure 5.27 The graph of z = / ext (jc, y). 




5.2 | Double Integrals 





jy = 3x 2 




Djl(l,3) 




/ \ y = 4 - X 2 




(2,0) 



Figure 5.28 The domain of / of 
Example 5. 



DEFINITION 2.9 Under the previous assumptions and notation, if R is any 
rectangle that contains D, we define 



fL fdA ,obe //« 



/ ext JA. 



Note that Definition 2.9 implicitly assumes that the choice of the rectangle R 
that contains D does not affect the value of f f R f ext dA. This is almost obvious 
but still should be proved. We shall not do so directly but instead establish the 
following key result: 

THEOREM 2.10 Let D be an elementary region in R 2 and / a continuous 
function on D. 

1. If D is of type 1 (as described in Definition 2.8), then 

/ / fdA = / / f(x,y)dydx. 

J J D Ja Jy(x) 

2. If D is of type 2, then 

r r rd pP(y) 

fdA= / f(x,y)dxdy. 

J J D Jc Ja(y) 



Theorem 2.10 provides an explicit and straightforward way to evaluate double 
integrals over elementary regions using iterated integrals. Before we prove the 
theorem, let us illustrate its use. 

EXAMPLE 5 Let D be the region bounded by the parabolas y = 3x 2 , y = 
4 — x 2 and the y-axis as shown in Figure 5.28. (Note that the parabolas intersect at 
thepoint(l, 3).) Since D isatype 1 elementary region, we may use Theorem 2. 10 
with f(x, y) — x 2 y to find that 



2 f l r 4 ^ 2 2 

x ydA= II x ydydx. 
Jo J\k 2 



The limits for the first (inside) integration come from the y-values of the top and 
bottom boundary curves of D. The limits for second (outside) integration are the 
constant x -values that correspond to the straight left and right sides of D. The 
evaluation itself is fairly mechanical: 

=4-x 2 



I I x 2 ydydx = I I 

J0 Jxx 2 Jo V 



1/22 

' x y 



dx 



=3x 2 



= / y((4-* 2 ) 2 -(3* 2 )>* 
l f l 

= -J x 2 (16 - 8;t 2 + x 4 - 9x 4 ) dx 

= f (8.t 2 - 4x 4 - 4x 6 ) dx = 
Jo 



8 4 4 _ 136 
3 5 7 105 - 



324 Chapter 5 | Multiple Integration 



Note that after the y -integration and evaluation, what remains is a single definite 
integral in x. The result of calculating this x -integral is, of course, a number. Such 
a situation where the number of variables appearing in the integral decreases with 
each integration should always be the case. ♦ 



Proof of Theorem 2.10 For part 1, we may take D to be described as 

D = {(x, y) | y(x) < y < S(x), a < x <b}. 
We have, by Definition 2.9, that 



fL fdA =fL f ~ dA - 



where R is any rectangle containing D. Let R = [a', b'] x [c', d'], where a' < a, 
b' > b, and c' < y(x), d' > S(x) for all x in [a, b]. That is, we have the situation 
depicted in Figure 5.29. Since / ext is zero outside of the subrectangle R2 = 
[a,b] x [c',d'], 

f f f ext dA= f f f ext dA= f f f ex \x,y)dydx 

by Fubini's theorem. For a fixed value of x between a and b, consider the y- 
integral f ext (x, y)dy. Since f ext (x, y) = 0 unless y(x) < y < S(x) (in which 
case f ext (x, y) = f(x, y)), 



pel' pS(x) 

/ r x \x,y)dy= / f(x,y)dy, 

Jd Jy(x) 



and so 



f f f(x,y)dA= f f f xt dA= f f f ex \x,y)dydx 

J JD J J R Ja Jc' 



/ / f(x,y)dydx, 



as desired. 

The proof of part 2 is very similar. 




Figure 5.29 The region R is the union of Ri, R2, 
and^?3 . 



(0,1)' 






>v x + y = 1 




D 


(0,0) 


\(1.0) 


Figure 


5.30 The region D of 


Example 6. 


; 


> 




\ J = 1 - X | 


x = 0 






D 




y = 0 


Figure 


5.31 The region D of 


Example 6 as a type 1 region. 


; 


> 




y = l 


x = 0 


\x = 1 — y 




D \. 




y = 0 


Figure 


5.32 The region D of 



Example 6 as a type 2 region. 



5.2 | Double Integrals 325 

We continue analyzing examples of double integral calculations. 

EXAMPLE 6 Let D be the region shown in Figure 5.30 having a triangular 
border. Consider f f D (\ — x — y)dA. Note that D is a type 3 elementary region, 
so there should be two ways to evaluate the double integral. 

Considering D as a type 1 elementary region (see Figure 5.3 1), we may apply 
part 1 of Theorem 2.10 so that 

/ / (1 — x — y)dA= I I (1 — x — y)dydx 
J 3d Jo Jo 



-L 
-L 
-I 



y-xy 



y=\-x 



dx 



y=0 



' 7 (1 -x) 2 \ 
(1 — x) — x(l — x) I dx 



(1 - xf 



dx= -\{\-x)\ = \. 



We can also consider D as a type 2 elementary region, as shown in Figure 5.32. 
Then, using part 2 of Theorem 2.10, we obtain 

/ / (1 — x — y)dA = I I (l-x-y)dxdy. 

J J D Jo Jo 

We leave it to you to check explicitly that this iterated integral also has a value of 
t . Instead we note that 



Jo Jo 



y ) dy dx 



can be transformed into 



/ / (1 — x — y) dx dy 
Jo Jo 

by exchanging the roles of x and y. Hence, the two integrals must have the same 
value. In any case, the double integral 




(1 — x — y)dA 



represents the volume under the graph of z = 1 — x — y over the triangular region 
D . If we picture the situation in R 3 , as in Figure 5 . 3 3 , we see that the double integral 
represents the volume of a tetrahedron. ♦ 



Of course, not all regions in the plane are elementary, including even some 
relatively simple ones. To integrate continuous functions over such regions, the 
best advice is to attempt to subdivide the region into finitely many of elementary 
type. 



326 Chapter 5 | Multiple Integration 




EXAMPLE 7 Let D be the annular region between the two concentric circles 
of radii 1 and 2 shown in Figure 5.34. Then D is not an elementary region, but we 
can break D up into four subregions that are of elementary type. (See Figure 5.35.) 
If f(x, y) is any function of two variables that is continuous (hence integrable) 
on D, then we may compute the double integral as the sum of the integrals over 
the subregions. That is, 

[[ fdA=[[ fdA+ff fdA+ff fdA+[[ fdA. 

J J D J J Di J JD 2 J J D] J J D 4 

For the type 1 subregions, we have the set-up shown in Figure 5.36: 

f(x, y)dy dx 



and 



\L SiA 



V3 ,-1 



f(x, y)dy dx. 



-VI J—jA^? 

For the type 2 subregions, we use the set-up shown in Figure 5.37: 
f fo 2 f-i f*/i 




y = — V4 - x- 



Figure 5.36 The subregions D\ and 
Z?3 of Example 7 are of type 1 . 



fix, y)dx dy 


J 


1 


.2 _ 

\ ✓ 












\ 

\ 

s 

X 


✓ 



x= VT^y2 



x = V4-y 2 



Figure 5.37 The subregions D2 and D4 of 
Example 7 are of type 2. 



5.2 | Double Integrals 



and 



JL<«-U. 



<l-y 2 



-1 J-J4-V 2 



f(x, y)dx dy. 



The difficulty of evaluating each of the preceding four iterated integrals then 
depends on the complexity of the integrand. ♦ 



2 




/ D 


/x-y=0 


~K ~ l / 


1 2 


-l 





Figure 5.38 The region D 
of Example 8. 



EXAMPLE 8 We calculate ff D ydA, where D is the region bounded by the 
line x — y = 0 and the parabola x = y 2 — 2. (See Figure 5.38.) 

In this case D is a type 2 elementary region, where the left and right boundary 
curves may be expressed as x = y 2 — 2 and x = y, respectively. These curves 
intersect where 



y = -l,2. 



y*-2 = y y^-y-2 = Q < 

Therefore, part 2 of Theorem 2.10 applies to give 

-2 py 



1L 



ydA 



/I y dx dy 
■1 Jv 2 -2 



f i f 2 2 xy\^_ 2 dy = j y- (y 2 - 2)y) dy 
j y-y'+2y) dy- 



3 4 \ ^ 

y -- y - + y 2 
3 4 y 



= (f-4 + 4)-(- 



3 4 



+ !) = !■ 



Now, although D is not a type 1 elementary region, it may be divided into 
two type 1 subregions along the vertical line x = — 1. (See Figure 5.39.) The 
subregion D\ lying left of the line x = — 1 is bounded on both top and bottom by 
the parabola x = y 2 — 2; by solving for y we may express the bottom boundary 
of D\ as y = —*Jx + 2 and the top boundary as y = \fx + 2. The subregion 
D2 lying right of x = — 1 is bounded on the bottom by y = x and on the top by 




Figure 5.39 The region D of Example 8 is divided 
into subregions D\ and D2 by the line x = — 1. 



328 Chapter 5 | Multiple Integration 



y = -J x + 2. Putting all this information together, we have 
/ / ydA = I I ydA+ f f ydA 



/ — 1 p*/x+2 r2 r 

I _ydydx+ / / 
-2 J-Jx+2 J -I Jx 



y dy dx 



-1 y 2 

<-2 ~2 

-1 



y=Vx+2 



dx + 



£ 



2 2 y=vx+2 



y=-Vx+2 

/■ 2 /x + 2 * 2 \ 

- y ( — 



, x 3 
1 4 + *"T 



= (l+2-l)-(i-l + i) 



4' 



Addendum: Proof of Theorem 2.6 

Step 1. First we establish Theorem 2.6 in the case where / is continuous on 
R = [a, b] x [c, d]. By Theorem 2.4, we know that ff R f dA exists. Let F be 
the single-variable function denned by 



F(x) 



f(x,y)dy. 



(Note: Since / is continuous on R, the partial function in y obtained by holding 
x constant is continuous on [c, d]. Hence, f d fix, y)dy exists for every x in 
[a, b].) We show that 

nb nb r nd n n 

/ F(x)dx= / / f(x,y)dy dx = f dA. 

J a J a \_Jc J J JR 

Let a = xq < x\ < • • • < x„ = b be any partition of [a, b]. Then a general 
Riemann sum that approximates F(x) dx is 

// 

j>(jtf)A*,, (1) 



where Ax; = x ; - — andx* e [x,-_i, x,].Nowletc = y 0 < < • • • < = d 
be a partition of [c, d]. (The partitions of [a, b] and [c, J] together give a partition 
of R = [a, b] x [c, rf].) Therefore, we may write 



x) = J d f(x, 



y)dy 



fix, y)dy + 



y)dy + 



t f 

7=1 J y>- 1 



f(x,y)dy. 



+ f fix,y)dy 

Jy„-i 



5.2 | Double Integrals 329 

By the mean value theorem for integrals, 1 on each subinterval [yj-i, yj] there 
exists a number y* such that 



f 

Jy, 



f(x, y)dy = ( yj - y } -i)f(x, y*) = f(x, y*)A yj . 



The choice of y* in general depends on x, so henceforth, we will write y*(x) for 
y*. Consequently, 

n 

/•(.v)- ^./(.v.y;(.v)).\y,. 



and the Riemann sum (1) may be written as 



i=l I 7=1 



Ax i = /(C ; ,).\.V,.\.V ( . 



where cy = (xf , y*(xf)). Note that Cy e x ; ] x [y ; _i, y,-]. (See Fig- 
ure 5.40.) 



y;(*,*) 

































c • 











































H 1 — h 



H h 



1 « I 



Figure 5.40 The point Cy = (.«*, y *(*,*)) used in 
the proof of Theorem 2.6. 

We have thus shown that given any partition of [a,b], we can associate a 
suitable partition of R = [a, b] x [c, d] such that the Riemann sum (1) that ap- 
proximates f a F(x)dx is equal to a Riemann sum (namely, J2i j f( c ij)Axj Ayj) 
that approximates ff R f dA. Since / is continuous, we know that 

/(cy)Ajc,- Ay 7 - approaches / / f dA 
U J Js 

as Ax, and Ay,- tend to zero. Hence, 

I F(x)dx = 11 fdA. 

Ja J JR 

By exchanging the roles of x and y in the foregoing argument, we can show 

that 

II fdA = f I f(x,y)dxdy. 

J J R Jc Ja 



1 The mean value theorem for integrals says that, if g is continuous on [a, b], then there is some number 
c with a < c < b such that f g(x)dx = (b — a)g(c). 



Chapter 5 | Multiple Integration 



Step 2. Now we prove the general case of Theorem 2.6 (i.e., the case that 
/ has discontinuities in R = [a, b] x [c, d]). By hypothesis, the set S of discon- 
tinuities of / in R are such that every vertical line meets S in at most finitely 
many points. Thus, for each x in [a, b], the partial function in y of f(x, y) is 
continuous throughout [c, d], except possibly at finitely many points. (In other 
words, the partial function is piecewise continuous.) Then, because / is bounded, 



(x) = ^ f(x 



F(x)= I f(x,y)dy 
exists. 

Now we proceed as in Step 1 . That is, we begin with a partition of [a, b] into 
n subintervals and a corresponding Riemann sum 

i=i 

Next, we partition [c, d] into n subintervals. Hence, 

Fix*) = f f(x*, y)dy = J2 f' fix*, y)dy. (2) 

As in Step 1, the partitions of [a, b] and [c, d] combine to give a partition of R. 
Write R as U R 2 , where R\ is the union of all subrectangles 

Rij = [Xi-uXf] x [y ; _!, yj] 

that intersect S and R2 is the union of the remaining subrectangles. Then we may 
apply the mean value theorem for integrals to those intervals [yj-i, yj] on which 
fix*, y) is continuous in y, thus replacing the integral 

/ f(xf,y)dy 

by 

f(x*,y*(x*))Ay j = f(c i j)Ay j . 

Since / is bounded, we know that 

\f(x,y)\<M 

for some M and all (x, y) e R. Therefore, on the intervals [y 7 _i, yj] where 
fi x * , y) fails to be continuous, we have 

f 1 fix*,y)dy < r \ f(x*,y)\dy 

< r Mdy = M(yj - y ; _j) = MAy r (3) 

Jyj-i 

From equation (2), we know that 

f2Hx*)Axi = it,\f } fix*,y)dy}A Xi 

= E f fix*,y)dy\A Xi 



5.2 | Double Integrals 331 



f(x*,y)dy\Axi 



f(x*,y)dy\Ax 



Therefore, 



= s \i: 

A',,t K- 1 *' '• • 

J>(x*)Ax,- J] { r f(xf,y)dy\Ax i 

- (f 



f(x*,y)dy^Axi 



(4) 



Applying the mean value theorem for integrals to the left side of equation (4) and 
inequality (3) to the right side, we obtain 



J2nx*)Axi- f(Cij)AxtAyj 

i = l RijCRi 



MAxjAyj 

RijCRi 

M ■ area of R\ . 



Now S has zero area (by hypothesis) and is contained in R\. By letting the 
partition of R become sufficiently fine (i.e., by making Ax, , Ay, small), the term 
M ■ area of R\ can be made arbitrarily small. (See Figure 5.41.) 



d - 

c - 


















































- — 


























N 






< 






















f 












































i 




















































— - 








































































1 

a 


b 



Figure 5.41 The set R\ (shaded area) consists of 
the subrectangles of the partition of R that meet S, 
the set of discontinuities of / on R. As the partition 
becomes finer, the area of Ri tends toward zero. 



Therefore, as all Ax, and Ay,- tend to zero, we have that the sums 
J2F(xf)Axi and /<e»/).Yv,.\.\> 

i RijCRi 

and the term M ■ area of R\ converge (respectively) to 

f F(x)dx, f f fdA, and 0. 

J a J JR 



332 Chapter 5 | Multiple Integration 

We conclude that 

that is, 



5.2 Exercises 



f F(x)dx - f f fdA = 0, 

J a J J R 

( ( fdA = f f f(x,y)dydx. 

J J R J a Jc 

Again, by exchanging the roles of x and y, we can show that 

"d pb 



as well. 



[ ( fdA = f f f(x,y)dxdy 

J JR Jc J a 



1. Use Definition 2.3 and Theorem 2.4 to determine 
the value of f f R (y 3 + s'm2y)dA, where R = [0, 3] x 
[-1,1]- 

2. Let R = [-3, 3] x [-2, 2]. Without explicitly eval- 
uating any iterated integrals, determine the value of 
Jf R (x 5 +2y)dA. 

3. This problem concerns the double integral ffpX^dA, 
where D is the region pictured in Figure 5.42. 

(a) Determine ff £) x i dA directly by setting up and ex- 
plicitly evaluating an appropriate iterated integral. 

(b) Now argue what the value of ff D x 3 dA must be 
by inspection, that is, without resorting to explicit 
calculation. 




Figure 5.42 The region D of 
Exercise 3 . 

In Exercises 4-13, evaluate the given iterated integrals. In ad- 
dition, sketch the regions D that are determined by the limits 
of integration. 



, f f 3d 

Jo Jo 



r 2 r 1 

i. I I y dx dy 

Jo Jo 



Jo Jo 



lo Jo 

^3 ?2x+\ 



y dy dx 



-1 xydy 



dx 



2 + y 2 )dy dx 



10. 



11. 



12. 



13. 



/ / {x 

Jo Jx 2 /4 
,4 ,2,/> 

/ / x sin iy ) dx dy 
Jo Jo 

pit psinx 

11 y cos x dy dx 
Jo Jo 

f f ' 3dydx 
Jo J-JT^x 1 

Li zdxd > 

lo J-t 



y dy dx 



14. Figure 5.43 shows the level curves indicating the vary- 
ing depth (in feet) of a 25 ft by 50 ft swimming pool. 
Use a Riemann sum to estimate, to the nearest 100 ft 3 , 
the volume of water that the pool contains. 




Figure 5.43 

1 5. Integrate the function f(x, y) = 1 — xy over the trian- 
gular region whose vertices are (0, 0), (2, 0), (0, 2). 



5.2 I Exercises 333 



16. Integrate the function f(x, y) = 3xy over the region 
bounded by y = 32x 3 and y = ~Jx. 

17. Integrate the function f(x, y) = x + y over the region 
bounded by x + y = 2 and y 2 — 2y — x = 0. 

18. Evaluate ff D xydA, where D is the region bounded 
by x = y 3 and y = x 2 . 

19. Evaluate ff D e* 2 d A, where D is the triangular region 
with vertices (0, 0), (1, 0), and (1, 1). 

20. Evaluate ff D 3yd A, where D is the region bounded 
by xy 2 = 1, y = x, x = 0, and y = 3. 

21. Evaluate ff D (x — 2y)dA, where D is the region 
bounded by y = x 2 + 2 and y = 2x 2 — 2. 

22. Evaluate Jf D (x 2 + y 2 )dA, where D is the region in the 
first quadrant bounded by y = x, y = 3x, and xy = 3. 

23. Prove property 2 of Proposition 2.7. 

24. Prove property 3 of Proposition 2.7. 

25. Prove property 4 of Proposition 2.7. 

26. (a) Let D be an elementary region in R 2 . Use the 

definition of the double integral to explain why 
ff D IdA gives the area of D. 

(b) Use part (a) to show that the area inside a circle of 
radius a isjia 2 . 

27. Use double integrals to find the area of the region 
bounded by y = x 2 and y = x 3 . 

28. Use double integrals to calculate the area of the region 
bounded by y = 2x, x = 0, and y = 1 — 2x — x 2 . 

29. Use double integrals to calculate the area inside 
the ellipse whose semiaxes have lengths a and b. 
(See Figure 5.44.) 




Figure 5.44 The ellipse of 
Exercise 29. 



30. (a) Set up an appropriate iterated integral to find 
the area of the region bounded by the graphs of 
y = — x and y = ax 2 for x > 0. (Take a to be 
a constant.) 

(b) Use a computer algebra system to estimate for what 
value of a this area equals 1 . 



31 . Use double integrals to find the total area of the region 
bounded by y = x 3 and x 



y 5 - 



32. Use double integrals to find the area of the region 
bounded by the parabola y = 2 — x 2 , and the lines 
X - y = 0, 2x + y = 0. 

33. Let D be the region in R 2 bounded by the lines x = 0, 
x + y = 3, and x — y = 3. Without resorting to any 
explicit calculation of an iterated integral, determine, 
with explanation, the value of ff D (y 3 — e x siny + 
2)dA. (Hint: Use symmetry and geometry.) 

34. Let D be the region in R 2 with y > 0 that is 
bounded by x 2 + y 2 = 9 and the line y = 0. Without 
resorting to any explicit calculation of an iterated 
integral, determine, with explanation, the value of 



ff D V* 



y sin* — 2) dA. 



35. Determine the volume of the solid lying under the plane 
z = 24 — 2x — 6y and over the region in the jcy-plane 
bounded by y = 4 — x 2 , y = 4x — x 2 , and the y-axis. 



36. Find the volume under the portion of the paraboloid 
z = x 2 + 6y 2 lying over the region in the jcy-plane 
bounded by y = x and y = x 2 — x. 

37. Find the volume under the plane z = 4x + 2y + 25 
and over the region in the xy-plane bounded by y = 
x 2 - 10 and y = 31 - (x - \ f. 

38. (a) Set up an iterated integral to compute the volume 

under the hyperbolic paraboloid z = x 2 — y 2 + 5 
and over the disk 



D = {(x, y) 



x 2 + y 2 < 4} 



in the xy -plane. 

(b) Use a computer algebra system to evaluate the 
integral. 

39. Find the volume of the region under the graph of 

f(x,y) = 2-\x\-\y\ 
and above the xy-plane. 

40. (a) Show that if R = [a, b] x [c, d], f is continuous 

on [a, b], and g is continuous on [c, d], then 



f(x)g(y)dA 



{f, md ') if.™") 



(b) What can you say about 

f{x)g(y)dA 



11 



if D is not a rectangle? More specifically, what if 
D is an elementary region of type 1 ? 



334 Chapter 5 Multiple Integration 



41. Let 



/(*. y) 



1 if x is rational 

0 if x is irrational and y < 1 

2 if x is irrational and y > 1 



(a) Show that f Q f(x,y)dy does not depend on 
whether x is rational or irrational. 



(b) Show that 
value. 



fo f( x > y)dy dx exists and find its 



(c) Partition R = [0, 1] x [0, 2] and construct a 
Riemann sum by choosing "test points" Cy in each 
subrectangle of the partition to have rational x- 



coordinates. Then to what value must this Riemann 
sum converge as both Ax; and Ay j tend to zero? 

( d) Partition R and construct a Riemann sum by choos- 
ing test points Cy = (xf, y *) such that x* is rational 
if y] < 1 and xf is irrational if y * > 1 . What hap- 
pens to this Riemann sum as both A.t, and At- 
tend to zero? 

(e) Show that / fails to be integrable on R by us- 
ing Definition 2.3. Thus, we see that double inte- 
grals and iterated integrals are actually different 
notions. 



5.3 Changing the Order of Integration 

Frequently, it is useful to think about the evaluation of double integrals over 
elementary regions essentially as the determination of an appropriate order of 
integration. When the region of integration is a rectangle, Fubini's theorem 
(Theorem 2.6) says the order in which we integrate has no significance; that is, 

fffdA=f f f(x,y)dydx=f f f(x,y)dxdy. 

J J R J a J c J c J a 

(See Figure 5.45.) When the region is elementary of type 1 only, we must integrate 
first with respect to y (and then with respect to x) if we wish to evaluate the double 
integral by means of a single iterated integral. (See Figure 5.46.) Then 



fdA= / f(x,y)dydx 

J JD J a Jy(x) 



In the same way, when the region is elementary of type 2 only, we would typically 
integrate first with respect to x, so that 



r r rd rP(y) 

fdA= / f(x,y)dxdy. 

J JD Jc Ja(y) 











® 


® 





































® 


® 





























R = [a, b] x [c, d] 



x = a 


y = 8(x) 
® 




® 


x = b 








y=y(x) 









Figure 5.45 Changing the order of integration over a 
rectangle. 



Figure 5.46 A type 1 region 
leads us to integrate with respect 
to y first. 



5.3 | Changing the Order of Integration 335 




Figure 5.47 A type 2 region 
leads to integration with respect 
to x first. 




Figure 5.48 The region D of 
Example 1. 




Figure 5.49 Integrating over 
the region D of Example 1 by 
integrating first with respect to x. 



(See Figure 5.47.) When the region is elementary of type 3, however, we can 
choose either order of integration, at least in principle. Often, this flexibility can 
be used to advantage, as the following examples illustrate: 

EXAMPLE 1 We calculate the area of the region shown in Figure 5.48. Con- 
sidering D as a type 1 region, we obtain 



Area of 



p p p e p m X 

D= Id A (Why?) = / / 1 dy dx 

J J D J\ JO 

y | Q x dx = J In x dx . 



The single definite integral that results gives the area under the graph of y = In x 
over the ^-interval [1, e], just as it should. To evaluate this integral, we need to 
use integration by parts: Let u = In x (so du = 1 /x dx) and dv = dx (so v = x). 
Then 



Area of D 



[ 



In x dx = In x ■ x\ 



(VI 



dx 



(remember f u dv = u ■ v — f v du), so 
Area of D = e — 0 



j dx = e — (e — 1) 



1. 



Integration by parts can be avoided if we integrate first with respect to x, as 
schematically suggested by Figure 5.49. Hence, 



Area of D 



I I \dA = I I \dxdy= I x\ e y dy= I (e — e y )dy 
J Jd Jo Je? Jo e Jo 



= (ey-e?)\ l 0 = (e-e)-(0-e°)=l, 
which checks (just as it should). ♦ 

Note that the two iterated integrals we used to calculate the area in Example 1 , 
namely, 

/» e p In x f^f e 
II dy dx and / / dxdy, 
J i Jo Jo Jey 

are not obtained from each other by a simple exchange of the limits of integration. 
The only time such an exchange is justified is when the region of integration 
is a rectangle of the form [a, b] x [c, d] so that all limits of integration are 
constants. 

EXAMPLE 2 Sometimes changing the order of integration can make an im- 
possible calculation possible. Consider the evaluation of the following iterated 
integral: 



LI 



y cos(x 2 )dx dy. 



After some effort (and maybe some scratchwork), you should find it impossible 
even to begin this calculation. In fact it can be shown that cos(x 2 ) does not 
have an antiderivative that can be expressed in terms of elementary functions. 
Consequently, we appear to be stuck. 



Chapter 5 | Multiple Integration 




Figure 5.50 Note that x 



corresponds to y 
region shown. 



fx over the 



On the other hand, it is easy to integrate y cos(x 2 ) with respect to y. This 
suggests finding a way to change the order of integration. We do so in two steps: 

1. Use the limits of integration in the original iterated integral to identify the 
region D in R 2 over which the integration takes place. (While doing this, you 
should make a wish that D turns out to be a type 3 region.) 

2. Assuming that the region D in Step 1 is of type 3, change the order of 
integration. 

The limits of integration in the preceding example imply that D can be described as 

D = {(x, y) | v 2 < x < 4, 0 < y < 2}, 

as suggested by Figure 5.50. Now Figure 5.50 can be used to change the order 
of integration. We have 



a: 



ycos(x )dxdy 



ff 

Jo Jo 



y cos(x )dy dx. 



It is now possible to complete the calculation; that is, 



ff 

Jo Jo 



y cos(x 2 ) dy dx 



-f 

-jfi 



— cos(x 2 ) 



cos(x ) dx 



dx 



y=0 



16 



cos u du, 



where u = x and du = 2x dx, so that, finally, 

-2 



if 



ycos(x )dxdy 



1 • 1 16 

4 smM lo 



sin 16. 



The technique of changing the order of integration is a very powerful one, but 
it is by no means a panacea for all cumbersome (or impossible) integrals. It relies 
on an appropriate interaction between the integrals and the region of integration 
that often fails to occur in practice. 



5.3 Exercises 



1 . Consider the integral 

r2x 



and check that your answer agrees with part (a). 

In Exercises 2-9, sketch the region of integration, reverse the 
order of integration, and evaluate both iterated integrals. 



'-ff 

Jo Jo 



(2 — x — y ) dy dx 



j j {2x+\)dydx. r 2 r*- 2 * 

Jo Jo 



y dy dx 



(a) Evaluate this integral. „2 m-/ 

(b) Sketch the region of integration. I I x dx dy 

(c) Write an equivalent iterated integral with the order 
of integration reversed. Evaluate this new integral 5. / / (x + y)dxdy 
and check that your answer agrees with part (a). Jo Jjy 



i. j j Idydx 



5.4 | Triple Integrals 



7. J J e x dxdy 

p7t/2 fCOSX 

i. I I sinx dydx 
Jo Jo 



' a: 



/4-y 2 



y dx dy 



/4-v 2 



When you reverse the order of integration in Exercises 10 and 
11, you should obtain a sum of iterated integrals. Make the 
reversals and evaluate. 

10. f [ (x-y)dvdx 
J -2 Jx 2 -2 



4 i-Ay-y 1 



11 



u 



(y + \)dxdy 



In Exercises 12 and 13, rewrite the given sum of iterated in- 
tegrals as a single iterated integral by reversing the order of 
integration, and evaluate. 



12 



r- 1 nx p2 p2—x 

. I /sin x dy dx + I I sin x dy dx 
Jo Jo Ji Jo 



13. 



/*/ 

Jo Jo 



8 rJyft yl2 i-^/y/i 

y dx dy + I I y dx dy 
Js "'Vy=S 



In Exercises 14—18, evaluate the given iterated integral. 



•/'/ 

Jo Jiy 
lo ly 

■a 



cos (x ) dx dy 



x sin xy dx dy 



smx 



dx dy 



■ IT' 

Jo Jo 

■ ff • 

JO Jy/2 



dy dx 



dx dy 



It is interesting to see what a computer algebra system does with 
iterated integrals that are difficult or impossible to integrate in 
the order given. In Exercises 19-21, experiment with a com- 
puter to evaluate the given integrals. 

19. (a) Determine the value of J fl 2 J^ 2 y 2 cos (xy) dy dx 

via computer. Note how long the computer takes 
to deliver the answer. Does the computer give you 
a useful answer? 

(b) If you were to calculate the iterated integral in part 
(a) by hand, in the order it is written, what method 
of integration would you use? (Don't actually carry 
out the evaluation, just think about how you would 
accomplish it.) 

(c) Now reverse the order of integration and let your 
computer evaluate this iterated integral. Does your 
computer supply the answer more quickly than in 
part (a)? 

20. (a) See if your computer can calculate 

Jo fx 2 x sm (y 2 ) dy dx as it is written, 
(b) Now reverse the order of integration and have 
your computer evaluate your new iterated integral. 
Which of the computations in parts (a) or (b) is 
easier for your computer? 

^^21. (a) Can your computer evaluate /J f^-i e cosx dx dy? 

(b) Reverse the order of integration and have it try 
again. What happens? 




f— y 



Figure 5.51 The box 

B = [a, b] x [c, d] x [p, q]. 



5.4 Triple Integrals 

Let f{x, y, z) be a function of three variables. Analogous to the double integral, 
we define the triple integral of / over a solid region in space to be the limit of 
appropriate Riemann sums. We begin by denning this integral over box-shaped 
regions and then proceed to define the integral over more general solid regions. 

The Integral over a Box 

Let B be a closed box in R 3 whose faces are parallel to the coordinate planes. 
That is, 

B = {{x, y, z) eR 3 | a < x < b, c < y <d, p < z <q}. 
(See Figure 5.51.) We also use the following shorthand notation for B: 
B = [a, b] x [c, d] x [p, q]. 



Chapter 5 | Multiple Integration 



b=x„ 



c = y 0 y\ yi 
y 



q = Z n 



P = Zo 



d = y n 



Figure 5.52 A partitioned box. 



DEFINITION 4.1 A partition of B of order n consists of three collections 
of partition points that break up B into a union of n 3 subboxes. That is, for 
i, j, k = 0, . . . , n, we introduce the collections {jc, }, {yj}, and {zk}, such that 

a = xo < x\ < ■ ■ ■ < < Xj < ■ • • < x„ = b, 

c = yo < y\ < ■ ■ ■ < yj-i < yj < ■■■ < y n = d, 

p = Zo < Z\ < ■ ■ ■ < Zk-l < Zk < ■ ■ ■ < z„ = q. 
(See Figure 5.52.) In addition, for i, j,k = 1 n, let 

Axj=Xi—Xi-u Ayj = yj — yj-\, and Azk = Zk - Zk-u 



DEFINITION 4.2 Let / be any function defined on B = [a, b] x [c, d] x 
[p, q]. Partition B in some way. Let Cy* be any point in the subbox 

Bijk = [xt-i,Xi] x [yj-uyj] x [z*-i. Zifc] (},j,k= I,..., n). 
Then the quantity 

n 

S= f(c ljk )AV ijk , 

i,j,k=l 

where AVfjk = AxjAyjAzk is the volume of B^, is called the Riemann 
sum of j f on B corresponding to the partition. 



You can think of the Riemann sum ^ f(Cijk)AVijk as a weighted sum of 
volumes of subboxes of B, the weighting given by the value of the function / at 
particular "test points" Cy£ in each subbox. 



DEFINITION 4.3 The triple integral of / on £, denoted by 

///.'"• 

by jjj f(x,y,z)dV, orby jjj f(x,y,z)dxdydz, 



5.4 | Triple Integrals 





/ 




o 




/ 

/ 

/ 







B 

Figure 5.53 The subbox 
contributes /(c,/a-)A V ; ,^ 
to the Riemann sum S. If 
we think of / as 
representing a density 
function, then the total 
mass of the entire box B 
KffhfdV. 



is the limit of the Riemann sum S as the dimensions Ax, , Ay, , and Azk of 
the subboxes Bjj k all approach zero, that is, 

fff fdV= lim T /(c, ;A )A v, Ay, A- A . 

J J J B allA«,A^,A«-.0,j^ 1 

provided that this limit exists. When fff B fdV exists, we say that / is 
integrable on B. 



The key point to remember is that the triple integral is the limit of Riemann 
sums. It is this notion that enables useful and important applications of integrals. 
For example, if we view the integrand / as a type of generalized density function 
("generalized" because we allow negative density!), then the Riemann sum S is 
a sum of approximate masses (densities times volumes) of subboxes of B. These 
approximations should improve as the subboxes become smaller and smaller. 
Hence, we can use the triple integral fff g f dV, when it exists, to compute the 
total mass of a solid box B whose density varies according to /, as suggested by 
Figure 5.53. 

Analogous to Theorem 2.5, we have the following result regarding integra- 
bility of functions: 

THEOREM 4.4 If / is bounded on B and the set of discontinuities of / on B 
has zero volume, then fff B fdV exists. (See Figure 5.54.) 




Figure 5.54 In 

Theorem 4.4 the 
discontinuities of / on B 
(shown shaded) must have 
zero volume. 



To say that a set X has zero volume as we do in Theorem 4.4, we mean that 

we can cover X with boxes B\, B 2 B n , . . . (i.e., so that X c [J™ =1 B n ), the 

sum of whose volumes can be made arbitrarily small. 

To evaluate a triple integral over a box, we can use a three-dimensional version 
of Fubini's theorem. 



THEOREM 4.5 (Fubini's theorem) Let / be bounded on B = [a, b] x 
[c, d] x [p, q] and assume that the set S of discontinuities of / has zero vol- 
ume. If every line parallel to the coordinate axes meets S in at most finitely many 
points, then 



///. 



fdV 



pi? pel pq pb pq pel 

— f{x,y,z)dzdydx= III 

J a J c J p J a J p J c 

pd pb pq pd I pq pb 

= / / / f(x,y,z)dzdxdy= / / / f(x,y,z)dxdzdy 

J c J a J p J c J p J a 



f(x, y, z)dy dz dx 



pq pb pd pq pd pb 

= I I I f(x,y,z)dydxdz= III f(x,y,z)dxdydz. 

J p J a J c J p J c J a 



q pd pb 



340 Chapter 5 | Multiple Integration 




Figure 5.55 The function / is 
continuous on W. 



z=Y(x,y) 




z = (p(x,y) 



-y 



Figure 5.56 An elementary 
region of type 1 . 





y = d 


x = a(y){^ 






y = c 


Figure 5.57 


The "shadow" 



(projection) of W into the .rj-plane 
should be an elementary region in 
the plane. 



EXAMPLE 1 Let 

B = [-2, 3] x [0, 1] x [0, 5], and let f(x, y, z) = x 2 e y + xyz. 

Thus, / is continuous and hence certainly satisfies the hypotheses of Fubini's 
theorem. Therefore, 

f f [ (xV +xyz)dV = f f f (x 2 e y +xyz)dzdydx 

J J Jb J -2 Jo JO 

J (x 2 e y z + \xyz 2 )\l =0 dy dx 
= j j (5x 2 e y + f xy) dy dx 
= ^ {5x 2 e y + 2 ixy%T^dx 

= j (5(e- l)x 2 + fx)dx 

= (f (e -iy + f^)| 3 _ 2 

= (45( e -l)+f)-(-f( e -l)+f) 

= f(,-D+f. 

You can check that integrating in any of the other five possible orders produces 
the same result. ♦ 



Elementary Regions in Space 

Now suppose W denotes a fairly arbitrary solid region in space, like a rock or a 
slab of tofu. Suppose / is a continuous function defined on W, such as a mass 
density function. (See Figure 5.55.) Then the triple integral of / over W should 
give the total mass of W. As was the case with general double integrals, we need to 
find a way to properly define fff w f dV and to calculate it in practical situations. 
The course of action is much like before: We see how to calculate integrals over 
certain types of elementary regions and treat integrals over more general regions 
by subdividing them into regions of elementary type. 



DEFINITION 4.6 We say that W is an elementary region in space if it can 
be described as a subset of R 3 of one of the following four types: 

Type 1 (see Figures 5.56 and 5.57) 

(a) W = {(x, y, z) | <p(x, y) < z < f(x, y), y(x) < y < S(x), a <x <b], 

or 

(b) W = {(x, y, z) | <p(x, y)<z< fix, y), a(y) <x< B(y), c<y<d}. 



5.4 | Triple Integrals 341 



Type 2 (see Figure 5.58) 

(a) W = {(x, y, z) | a(y, z) < x < P(y, z), y(z) <y< S(z), p<z<q), 

or 

(b) W = {(x, y, z) | a(y, z) < x < p(y, z), <p(y) <z< f(y), c<y< d}. 
Type 3 (see Figure 5.59) 

(a) W = {(x, y, z) | y(x, z) <y < S(x, z), a(z) <x< fi(z), p < z < q}, 

or 

(b) W = {(x, y, z) | y{x, z)< y < S(x, z), (p(x) <z< ir(x), a<x<b}. 



Type 4 



W is of all three previously described types. 



x= a(y,z) 




y=y(x,z) 



y = 5(x, z) 




Figure 5.58 For an elementary 
region of type 2, the shadow in the 
jz-plane should be an elementary 
region in the plane. 



Figure 5.59 For an elementary 
region of type 3, the shadow in the 
-tz-plane should be an elementary 
region in the plane. 



Some explanation regarding Definition 4.6 is in order. An elementary region 
W of type 1 is a solid shape whose top and bottom boundary surfaces each can 
be described with equations that give z as functions of x and y and such that 
the projection of W into the xy-plane (the "shadow") is in turn an elementary 
region in R 2 (in the sense of Definition 2.8). Similarly, an elementary region of 
type 2 is one whose front and back boundary surfaces each can be described with 
equations giving x as functions of y and z and whose projection into the yz-plane 
is an elementary region in R 2 . Finally, an elementary region of type 3 is one whose 
left and right boundary surfaces each can be described with equations giving y 
as functions of x and z and whose projection into the xz-plane is an elementary 
region in R 2 . In each case, an elementary region in space is one for which we 
can find boundary surfaces described by equations where one of the variables is 
expressed in terms of the other two, and whose "shadow" in the plane of these 
two variables is an elementary region in R 2 in the sense of Definition 2.8. 

EXAMPLE 2 Let W be the solid region bounded by the hemisphere x 2 + y 2 + 
z 2 = 4, where z < 0, and the paraboloid z = 4 — x 2 — y 2 . (See Figure 5.60.) It 
is an elementary region of type 1 since we may describe it as 



W = {(x, y, z) | -x 2 -y 2 < z < 4 

-V4 - x 2 < y < ^4 - x 2 , -2 < x < 2j 



2 2 
x - y , 



342 Chapter 5 | Multiple Integration 



z 




x 2 + y 2 + z 2 = 4, 

z < 0 Shadow of W 

Figure 5.60 The solid region W of Example 2. 



This description was obtained by noting that W is bounded on top and bottom by a 
pair of surfaces, each of which is the graph of a function of the form z = g(x, y) 
and the shadow of W in the xy-plane is a disk D of radius 2, which we have 
chosen to describe as 



D=\(x,y) | -J*- 



< y < 



-2 < x < 2 



and which we already know is an elementary region (of type 3) in the xy- 
plane. ♦ 

EXAMPLE 3 The solid bounded by the ellipsoid 

a,b,c positive constants 



2 2 2 

^ X L V Z 

E : — + J — + — 



b 2 



1. 



can be seen to be an elementary region of type 4. To see that it is of type 1 , split 
the boundary surface in half via the z = 0 plane as shown in Figure 5.61. (This 
is accomplished analytically by solving for z in the equation for the ellipsoid.) 
Then the shadow D of E is the region inside the ellipse in the xy-plane shown in 
Figure 5.62. 




Figure 5.61 The ellipsoid of 
Example 3 as an elementary 
region of type 1 . 




Figure 5.62 The shadow of 
the type 1 ellipsoid in 
Figure 5.61 is the region inside 
the ellipse x 2 /a 2 + y 2 /b 2 = 1 
in the xy-plane. 



5.4 | Triple Integrals 343 




Figure 5.63 The ellipsoid of 
Example 3 as a type 2 elementary 
region. 




Figure 5.64 The shadow 
of the ellipsoid in 
Figure 5.63 is the region 
inside the ellipse 
y 2 /b 2 + z 2 /c 2 = 1 in the 
yz-plane. 



We have 

D = \(x, y) 




Figure 5.65 The ellipsoid of 
Example 3 as a type 3 region. 



-bJ\ 



y < bJi 



-a < x < a 



= (*, y) 



-a 



'l-^, -b< y< b\ 




Figure 5.66 The 

shadow of the 
ellipsoid in 
Figure 5.65 in the 
Az-plane. 



so D is in fact a type 3 elementary region in R 2 . 

To see that E is of type 2, split the boundary at the x = 0 plane as in 
Figure 5.63. The shadow in the yz-plane is again the region inside an ellipse. 
(See Figure 5.64.) Finally, to see that E is of type 3, split along y = 0. (See 
Figures 5.65 and 5.66.) ♦ 

Triple Integrals in General — 

Suppose W is an elementary region in R 3 and / is a continuous function on W. 
Then, just as in the case of double integrals, we define the extension of / by 



.f(x,y,z) if(x,y,z)eW 
f (x, y,z) = { 

0 if(x,y,z)£W 



By Theorem 4.4, f ext is integrable on any box B that contains W. Thus, we can 
make the following definition: 



DEFINITION 4.7 Under the assumptions that W is an elementary region 
and / is continuous on W, we define the triple integral 

fff fdV tobe fff r x dV, 

J J Jw J J J B 

where B is any box containing W. 



344 Chapter 5 | Multiple Integration 



Using a proof analogous to that of Theorem 2.10, we can establish the 
following: 



THEOREM 4.8 Let W be an elementary region in R 3 and / a continuous 
function on W. 



1. If W is of type 1 (as described in Definition 4.6), then 



err fb r s ( x ) ri r Q c >y) 

/// fdV= / / f(x,y,z)dzdydx, (type la) 

J J JW J a Jy(x) J(fi(x,y) 



or 



r r r r d rP(y) rf( x >y) 

ill fdV= / / f(x,y,z)dzdxdy. (type lb) 

J J JW Jc Ja(y) J<p(x,y) 



2. If W is of type 2, then 



r r r rl r$(z) rP(y,z) 

fdV= / f(x,y,z)dxdydz, 

J J JW J p Jy(z) Ja(y,z.) 



or 



r r r rd rf(y) rP(y,z) 

// / fdV= / / f(x,y,z)dxdzdy. (type 2b) 

J J JW Jc Jtp(y) Ja(y,z.) 



3. If W is of type 3, then 



or 



r r r ri rP(z) rHx,z) 

fdV= / f(x,y,z)dydxdz, 

J J JW J p Ja(z) Jy(x,z) 



err r b rfU) rHx,z.) 

fdV= / / f(x,y,z)dydzdx. 

J J JW Ja Jtp(x) Jy(x,z) 



(type 2a) 



(type 3a) 



(type 3b) 



(0,0, 1). 



Plane x + y + z = 1 



.(0,1,0) 

-y 



(1,0, 0) 



Figure 5.67 The tetrahedron of 
Example 4. 



EXAMPLE 4 Let W denote the (solid) tetrahedron with vertices at (0, 0, 0), 
(1,0, 0), (0, 1, 0), and (0, 0, 1) as shown in Figure 5.67. Suppose that the mass 
density at a point (x, y, z) inside the tetrahedron varies as f(x, y, z) — 1 + xy. 
We will use a triple integral to find the total mass of the tetrahedron. 
The total mass M is 



HI fdV= fff (l+xy)dV. 
J J Jw J J Jw 



(See the remark before Theorem 4.4.) To evaluate this triple integral using iterated 
integrals, note that we can view the tetrahedron as a type 1 elementary region. 
(Actually, it is a type 4 region, but that will not matter.) The slanted face is given 
by the equation x + y + z = 1, which describes the plane that contains the three 
points (1, 0, 0), (0, 1, 0), and (0, 0, 1). Hence, by first integrating with respect to 



5.4 | Triple Integrals 345 



z and holding x and y constant, 



(0, 1, 0) , 


* Line x + y = 1 
\^ (in z = 0 plane) 




(1,0,0) 


Figure 5.68 The shadow in the 
xy-plane of the tetrahedron of 
Example 4 is a triangular region. 




Figure 5.69 The region W of 
Example 5. 



M= f f (f \\+xy)dz\dA 

J J shadow \J 0 / 

= 11 {l+xy){\-x-y)dA 

J J shadow 

= / / (l - x - y + xy - x 2 y - xy 2 ) dA. 

J J shadow 



The shadow of W in the x v-plane is just the triangular region shown in Figure 5.68. 
Thus, 



M = (l - x — y + xy - x 2 y - xy 2 ) dA 

J J shadow 

= / / (l — x — y + xy — x 2 y — xy 2 ) dy dx 
Jo Jo 

= f ((1 - je) - x(\ - x) - 1(1 - x) 2 + \x{\ - x f 
Jo 



10 

L 



l x 2 (l - xf - \x{\-xf)dx 



= G-f* + ^ 3 -|*V*=( 2 



1 _ 5^ 1 .3 _ l x 4) dx = (l x _ + l x 4 _ l_ x 5yL _ 1_ 



12' 



30 )\0 40" 



Note that M can also be written as a single iterated integral, namely, 



M 



pi p 1— x p 1 

Jo Jo Jo 



(1 + xy)dz dy dx. 



EXAMPLE 5 We calculate the volume of the solid W sitting in the first octant 
and bounded by the coordinate planes, the paraboloid z = x 2 + y 2 + 9, and the 
parabolic cylinder y = 4 — x 2 . (See Figure 5.69.) 

By definition, the triple integral is a limit of a weighted sum of volumes of 
tiny subboxes that fill out the region of integration. If the weights in the sum are 
all taken to be 1, then we obtain an approximation to the volume: 

i,j,k 



Therefore, taking the limit as the dimensions of the subboxes all approach zero, 
it makes sense to define 



v = ////""■ 



Chapter 5 Multiple Integration 



(0,4) 




Figure 5.70 The shadow 
in the xy-plane of the 
region in Figure 5.69. 




z = 3x 2 + 3y 2 -16 



gure 5.71 The capsule-shaped 
gion of Example 6. 




Figure 5.72 The shadow 
of the region W of 
Example 6, obtained by 
projecting the intersection 
curves of the defining 
paraboloids onto the 
xy-plane. 



In our situation, W is a type 1 region whose shadow in the xy-plane looks 
like the region shown in Figure 5.70. Thus, by Theorem 4.4, 



V = dV = / / dzdydx 

J J Jw Jo Jo Jo 

= / / (x 2 + y 2 + 9)dydx 
Jo Jo 

= f*(* l y + \y s + 9y\^)dx 

= f (x 2 (4 - x 2 ) + i(4 - x 2 ) 3 + 9(4 - x 2 )) dx 
Jo 

= f 2 (m-2lx 2 + 3x 4 - \x 6 )dx 



= (ip x _ 7 x 3 + fx 



J_ v7 )l - 



21 " 



35 



EXAMPLE 6 We find the volume inside the capsule bounded by the paraboloids 
z = 9 — x 2 — y 2 and z = 3x 2 + 3y 2 — 16. (See Figure 5.71.) 
Once again, we have 



V 



IdV, 



and again the region W of interest is elementary of type 1 . The shadow, or pro- 
jection, of W in the xy-plane is determined by 

{(x, y) e R 2 | there is some z such that (x, y, z) e W}. 

Physically, one can also imagine the shadow as the hole produced by allowing W 
to "fall through" the xy-plane. In other words, the shadow is the widest part of W 
perpendicular to the z-axis. From Figure 5.71, one can see that it is determined 
by the intersection of the two boundary paraboloids. The shadow itself is shown 
in Figure 5.72. The intersection may be obtained by equating the z-coordinates 
of the boundary paraboloids. Therefore, 



9 - x 2 - y 2 = 3x 2 + 3y 2 



Thus, by Theorem 4.4, 



16 



4x 2 + 4y 2 = 25 

* 2 + y 2 = 2 { = (l) 2 - 



f f f fS/2 r^/25/4-x 2 p9-x 2 -y 2 

V = dV = / / dzdydx 

J J Jw J -5/2 J-Jli/A-x 2 ^3jc 2 +3y 2 -16 



/•5/2 p^/25/4-x 2 p9-x 2 -y 2 

= 4 / / / dz dy dx. 

Jo Jo J3x 2 +3v 2 



v 2 -16 



This last iterated integral represents the volume of one quarter of the capsule. 
Hence, we multiply its value by 4 to obtain the total volume. The reason for this 



5.4 I Exercises 347 



manipulation is to make the subsequent calculations somewhat simpler (although 
the computation that follows is clearly best left to a computer). 
We compute 



V 



rS/2 *9-x*-f 

= 41 / / dzdydx 

J0 J0 Jix 2 +3y 2 -i6 



r 5/2 p^25/4-x 2 

= 4 / / (25 -4x 2 -4y 2 )dydx 

Jo Jo 



(f-x 2 ) 3/2 )jx 



= 4^ U25-4x% 

/•5/2 / 



2^ /25 _ x 2 _ 4 ^25 _ v 2\ 3 / 2 



25 _ r 2 _ 4 ^25 
4 A 3 



(f -x 2 ) 3/2 )jx 
(f -x 2 ) 3/2 )jx 



= 4 (4-f)(f-x 2 ) 3 ^x 
Jo 

f 5/2 

= ¥/ ( 2 i-x 2 f 2 d X . 
Jo 



Now let x = J sin#, so dx = | cos9 d9. Then 

32 f^ 2 /5 \ 3 5 1250 f^ 2 



32 f^ 2 /5 V 5 1250 Z^ 2 4 

V = — -cosS -cos6d8 = / cos 4 Ode 

3 Jo \2 / 2 3 Jo 

r n/2 ( 

L G 



1250 Z^ 2 /l \ 2 
— / (-(1+COS20) ^ 



625 /' 7r/2 , 

/ (1 + 2 cos 2(9 + cos 2 26) d6 

6 Jo 



625 
~6~ 



625 r 1 * ' \ 

(6» + sin26»)|o /2 + — / -(1 + cos 46) d6 

6 Jo 2 

12 (H = 



625 625 ( it \ 625n 

7t + 

12 



5.4 Exercises 



Evaluate the triple integrals given in Exercises 1—3. 



1. f [ [ xyzdV 

J J J[-l,l]x[0,2]x[l,3] 

2 SSL 



SSL 



(x 2 + y 2 + z 2 )dV 



3. I I dV 

/[l,e]x[l,e]x[l,e] 

4. Find the value of fff w zdV, where W = [—1, 2] x 
[2, 5] x [—3, 3], without resorting to explicit 
calculation. 



/[0,l]x[0,2]x[0,3] 



348 Chapter 5 | Multiple Integration 



Evaluate the iterated integrals given in Exercises 5—7. 

5. f I [ 3yz 2 dxdydz 
J-i Ji Jo 

6. J I I (x + 2y + z) dy dx dz 

J\ Jo Ji 

ply ry+z 
JO Jl+y Jz 



z dx dz dy 



8. (a) Let W be an elementary region in R 3 . Use the 

definition of the triple integral to explain why 
fff w 1 dV gives the volume of W. 

(b) Use part (a) to find the volume of the region 
W bounded by the surfaces z = x 2 + y 2 and z = 
9-x 2 -y 2 . 

9. Use triple integrals to verify that the volume of a ball 
of radius a is 4jra 3 /3. 

1 0. Use triple integrals to calculate the volume of a cone of 
radius r and height It . (You may wish to use a computer 
algebra system for the evaluation.) 

In Exercises 11—20, integrate the given function over the indi- 
cated region W. 

11. f(x, y, z) — 2x — y + z; W is the region bounded 
by the cylinder z = y 2 , the Jty-plane, and the planes 
x = 0, x = 1, y = —2, y = 2. 

12. f(x,y,z) = y; W is the region bounded by the 
plane x + y + z = 2, the cylinder x 2 + z 2 = 1, and 
y = 0. 

13. f(x,y,z) = Sxyz; W is the region bounded by 
the cylinder y = x 2 , the plane y + z = 9, and the 
xy-plane. 

14. f(x,y,z) = z", W is the region in the first octant 
bounded by the cylinder y 2 + z 2 = 9 and the planes 
y = x, x = 0, and z = 0. 

15. f(x, y,z) = 1 — z 2 ', W is the tetrahedron with vertices 
(0, 0, 0), (1, 0, 0), (0, 2, 0), and (0, 0, 3). 

16. f(x, y, z) = 3.x:; W is the region in the first octant 
bounded by z = x 2 + y 2 , x = 0, y = 0, and z = 4. 

17. f(x, y, z) = x + y; W is the region bounded by 
the cylinder x 2 + 3z 2 = 9 and the planes y = 0, 
x + y = 3. 

18. f(x, y, z) = z; W is the region bounded by z = 0, 
x 2 + Ay 2 = 4, and z = X + 2. 

19. f(x, y, z) = 4jc + y; W is the region bounded by x = 

y 2 , y = z, x = y, and z = 0. 

20. f(x, y, z) = x; W is the region in the first octant 



21. Find the volume of the solid bounded by z = 4 — x 2 , 
x + y = 2, and the coordinate planes. 

22. Find the volume of the solid bounded by the planes y = 
0, z = 0, 2y + z = 6, and the cylinder x 2 + y 2 = 9. 

23. Find the volume of the solid bounded by the paraboloid 
z = Ax 2 + y 2 and the cylinder y 2 + z = 2. 

24. Find the volume of the region inside both of the cylin- 
ders x 2 + y 2 = a 2 and x 2 + z 2 = a 2 . 

25. Consider the iterated integral 

• l-x 



lo 



f(x, y, z)dz dx dy. 



Sketch the region of integration and rewrite the inte- 
gral as an equivalent iterated integral in each of the five 
other orders of integration. 

26. Change the order of integration of 

f(x, y, z)dz dx dy 



fff 

Jo Jo Jo 



to give five other equivalent iterated integrals. 
27. Change the order of integration of 



fff 

Jo Jo Jo 



f(x, y, z)dz dy dx 



bounded by z = x + 2y , z = 6 
and y = 0. 



y , x 



o, 



to give five other equivalent iterated integrals. 

28. Consider the iterated integral 

/ / / 2dzdydx. 

Jo Jo Ay 2 

(a) This integral is equal to a triple integral over a solid 
region W in R 3 . Describe W. 

(b) Set up an equivalent iterated integral by integrating 
first with respect to z, then with respect to x, then 
with respect to y. Do not evaluate your answer. 

(c) Set up an equivalent iterated integral by integrating 
first with respect to y, then with respect to z, then 
with respect to x. Do not evaluate your answer. 

(d) Now consider integrating first with respect to y, 
then x, then z. Set up a sum of iterated integrals 
that, when evaluated, give the same result. Do not 
evaluate your answer. 

(e) Repeat part (d) for integration first with respect to 
x, then z, then y. 

29. Consider the iterated integral 

,2 n\sf4^ rA-y 1 

/ / / (x 3 +y 3 )dzdydx. 

3-2 Jo Jx 2 +3y 2 

(a) This integral is equal to a triple integral over a solid 
region W in R 3 . Describe W. 



5.5 I Change of Variables 349 



(b) Set up an equivalent iterated integral by integrating 
first with respect to z, then with respect to x, then 
with respect to y. Do not evaluate your answer. 

(c) Set up an equivalent iterated integral by integrating 
first with respect to x, then with respect to z, then 
with respect to y. Do not evaluate your answer. 



(d) Now consider integrating first with respect to x, 
then y, then z. Set up a sum of iterated integrals 
that, when evaluated, give the same result. Do not 
evaluate your answer. 

(e) Repeat part (d) for integration first with respect to 
y, then z, then x. 



5.5 Change of Variables 

As some of the examples in the previous sections suggest, the evaluation of a mul- 
tiple integral by means of iterated integrals can be a complicated process. Both the 
integrand and the region of integration can contribute computational difficulties. 
Our goal for this section is to see ways in which changes in coordinates can be 
used to transform iterated integrals into ones that are relatively straightforward to 
calculate. We begin by studying the coordinate transformations themselves and 
how such transformations affect the relevant integrals. 



Coordinate Transformations 

Let T: R 2 —> R 2 be a map of class C 1 that transforms the wu-plane into the xy- 
plane. We are interested particularly in how certain subsets D* of the wu-plane 
are distorted under T into subsets D of the xy-plane. (See Figure 5.73.) 



D* 

CO 



D = T(D*) 



Figure 5.73 The transformation T(m, v) = (x(u, v), y(u, v)) 
takes the subset D* in the uv-p\am to the subset D = {(x, y) | 
(x, y) = T(m, v) for some (u, ») e D*) of the xy-plane. 



EXAMPLE 1 Let T(u, v) = (u + 1 , v + 2); that is, let x = u + 1, y = v + 2. 
This transformation translates the origin in the Mu-plane to the point (1 , 2) in the 
xy-plane and shifts all other points accordingly. The unit square D* = [0, 1] x 
[0, 1 ], for example, is shifted one unit to the right and two units up but is otherwise 
unchanged as shown in Figure 5.74. Thus, the image of D* is D = [1,2] x [2,3]. 

♦ 

EXAMPLE 2 Let S(w, v) = (2m, 3d). The origin is left fixed, but S stretches all 
other points by a factor of two in the horizontal direction and by a factor of three 
in the vertical direction. (See Figure 5.75.) ♦ 

EXAMPLE 3 Composing the transformations in Examples 1 and 2, we obtain 

(T o S)(«, «) = T(2«, 3u) = (2m + 1, 3v + 2). 
Such a transformation must both stretch and translate as shown in Figure 5.76. ♦ 



350 Chapter 5 | Multiple Integration 



D 



D 



□ 

= T(D') 



Figure 5.74 The image of D* = [0, 1] x [0, 1] is 

D = [1, 2] x [2, 3] under the translation 
T(m, v) = (u + 1, v + 2) of Example 1. 



£>* = [0, 1] x [0, 1] 



D = S(D*) = [0, 2] x [0, 3] 



Figure 5.75 The transformation S of Example 2 is a scaling 
by a factor of 2 in the horizontal direction and 3 in the vertical 
direction. 



y 



ToS 



D 



D = [1, 3] x [2, 5] 



1 



Figure 5.76 Composition of the transformations of 
Examples 1 and 2. 



EXAMPLE 4 Let T(w, v) = (u + v, u — v). Because each of the component 
functions of T involves both variables u and v, it is less obvious how the unit 
square D* = [0, 1] x [0, 1] transforms. We can begin to get some idea of the 
geometry by seeing how T maps the edges of D* : 



Bottom edge: (u, 0), 0 < u < 1 

Top edge: (u, 1), 0 < u < 1 

Left edge: (0, v), 0 < v < 1 

Right edge: (1, v), 0 < v < 1 



T(m, 0) = (u, u); 
T(m, 1) = (u + 1, m — 1); 
T(0, v) = (v, -v); 
T(l, w) = O + l, 1 - v). 



By sketching the images of the edges, it is now plausible that the image of D* 
under T is as shown in Figure 5.77. ♦ 




Figure 5.77 The transformation T of Example 4. 



by 



5.5 | Change of Variables 351 
More generally, we consider linear transformations T: R 2 — > R 2 defined 

T(w, v) = (au + bv, cu + dv) 



a b 




u 


c d 




V 



where a, b, c, and d are constants. (Note: The vector (u, v) is identified with the 
u 

v 



2 x 1 matrix 



.) One general result is stated in the following proposition: 



PROPOSITION 5.1 Let A = 

fined by 



a b 
c d 



, where det A ^ 0. If T: R 2 -> R 2 is de- 



T(w, w) = A 



then T is one-one, onto, takes parallelograms to parallelograms and the vertices 
of parallelograms to vertices. (See §2. 1 to review the notions of one-one and onto 
functions.) Moreover, if D* is a parallelogram in the i/u-plane that is mapped 
onto the parallelogram D = T(D*) in the xy-plane, then 

Area of D = \ det A| ■ (Area of D*). 



EXAMPLE 5 We may write the transformation T(u, v) = (u + v, u 
Example 4 as 



v) in 



T(m, v) 



Note that 



det 



u 

v 



= -2^0. 



Hence, Proposition 5.1 tells us that the square D* = [0, 1] x [0, 1] must be 
mapped to a parallelogram D = T(Z)*) whose vertices are 

T(0, 0) = (0, 0), T(0, 1) = (1, -1), T(1,0) = (1,1), T(l, 1) = (2,0). 

Therefore, Figure 5 .77 is indeed correct and, in view of Proposition 5.1, could have 
been arrived at quite quickly. Also note that the area of D is | — 2| • 1 = 2. ♦ 

Proof of Proposition 5.1 First we show that T is one-one. So suppose 
T(w, v) = T(u', v'). We show that then u = u',v = v' . We have 

T(w, v) = T(u' , v') 

if and only if 

(au + bu , cu + dv) = (au' + bv', cu' + du'). 
By equating components and manipulating, we see this is equivalent to the system 



a(u — u') + b(v — v') = 0 
c(u - u') + d(v - v') = 0 



(1) 



352 Chapter 5 | Multiple Integration 



If a 7^ 0, then we may use the first equation to solve for u — u': 



u! = — (v- v') 



(2) 




Figure 5.78 The vertices of 
D* = {p + sn + tb | 0 < s,t < 
1 } are at p, p + a, p + b, 
p + a + b (i.e., where s and t take 
on the values 0 or 1). 




Figure 5.79 The image D of the 
parallelogram D* under the linear 
transformation T(u) = An. 



Hence, the second equation in (1) becomes 

be 



(v - v') + d(v -v') = 0 



or, equivalently, 



-be + ad 



(v - v') = 0. 



By hypothesis, det A = ad — be ^ 0. Thus, we must have v — v' = 0 and there- 
fore, u — u' = 0 by equation (2). If a = 0, then we must have both b 0 and 
c 0, since det A ^ 0. Consequently, the system (1) becomes 

J b(v -v') =0 
I c(u - u') + d(v - v') = 0 " 

The first equation implies v — v' = 0 and hence, the second becomes c(u — u') = 
0, which in turn implies u — u! = 0, as desired. 

To see that T is onto, we must show that, given any point (x, y) e R 2 , we can 
find (u, v) e R 2 such that T(m, v) = (x, y). This is equivalent to solving the pair 
of equations 

Iau + bv = x 
cu + dv = y 

for u and v. We leave it to you to check that 

dx — by ay — cx 

14 = ~~a — TT and v = ~1 — TT 
ad — be ad — be 

will work. 

Now, let D* be a parallelogram in the ww-plane. (See Figure 5.78.) Then D* 
may be described as 

D* = {u | u = p + + tb, 0 < s < 1, 0 < t < 1}. 

Hence, 

D = T(D*) = {Au ueD) 

= {A(p + 5a + rb) | 0 < s < 1, 0<?<1} 

= {Ap + ^Aa + rAb | 0 < s < 1, 0 < t < 1}. 

If we let p' = Ap, a' = Aa, and b' = Ab, then 

D = {p' + sa' + rb' | 0 < s < 1,0 < r < 1}. 

Thus, £> is also a parallelogram and moreover, the vertices of D correspond to 
those of D*. (See Figure 5.79.) 

Finally, note that the area of the parallelogram D* whose sides are parallel to 



a 2 



and b 



by 

b 2 



5.5 | Change of Variables 



may be computed as follows: 









j 


k 


Areaof/?* = ||axb|| = 


det 






0 






bi 


b 2 


0 



= \a.\b 2 — a 2 b\ 



Similarly, the area of D = T(D*) whose sides are parallel to 



a = 



and b' = 



b\ 
b' 



is 



a' 2 b\ j 







a 


b 




a\ 




aa\ - 


- ba2 


. a 2 . 




c 


d 




CL 2 




ca\ - 


\-da 2 


' b\ ' 




a 


b ~ 




" b x ' 




ab\ - 


Ybb 2 


, b 2_ 




c 


d 




. b2 . 




cb\ - 


Ydb 2 



Area of D = ||a' x b'|| = \a[b' 2 
Now, a' = Aa and b' = Ab. Therefore, 



and 



Hence, by appropriate substitution and algebra, 

Area of D = \(aai + ba2)(cb\ + db 2 ) — (ca\ + da-£)(ab\ + bb 2 )\ 

= \{ad — bc)(a\b 2 — a 2 b\)\ 

= I det A] • area of D*. 

Note that we have not precluded the possibility of D*'s being a "degenerate" 
parallelogram, that is, such that the adjacent sides are represented by vectors a 
and b, where b is a scalar multiple of a. When this happens, D will also be 
a degenerate parallelogram. The assumption that det A 7^ 0 guarantees that a 
nondegenerate parallelogram D* will be transformed into another nondegenerate 
parallelogram, although we have not proved this fact. ■ 

Essentially all of the preceding comments can be adapted to the three- 
dimensional case. We omit the formalism and, instead, briefly discuss an example. 

EXAMPLE 6 Let T: R 3 -> R 3 be given by 

T(m, v, w) = (2u, 2u + 3v + w, 3w). 

Then we rewrite T by using matrix multiplication: 



Note that if 





- 2 


0 


0 " 




u 


w) 


2 


3 


1 




V 




_ 0 


0 


3 _ 




w 




" 2 


0 


0 " 




\ = 


2 


3 


1 


» 




0 


0 


3 





then det A = 18^0. 



354 Chapter 5 | Multiple Integration 



A result analogous to Proposition 5.1 allows us to conclude that T is one-one 
and onto, and T maps parallelepipeds to parallelepipeds. In particular, the unit 
cube 

D* = [0, 1] x [0, 1] x [0, 1] 

is mapped onto some parallelepiped D = T(£>*) and, moreover, the volume of D 
must be 

| det A| - volume of D* = 18- 1 = 18. 
To determine D, we need only determine the images of the vertices of the cube: 



r(o, o, 0) = (o, o, o) 

T(0, 0, 1) = (0, 1,3) 
T(0, 1, 1) = (0,4, 3) 
Both D* and its image D are shown in Figure 5.80. 



7X1, 0, 0) = (2, 2, 0); T(0, 1,0) = (0, 3, 0); 
T(l, 1, 0) = (2, 5, 0); 7X1, 0, 1) = (2, 3, 3); 
J(l, 1, 1) = (2, 6, 3). 





D* 


























Figure 5.80 The cube D* and its image D under the 
linear transformation of Example 6. 

EXAMPLE 7 Of course, not all transformations are linear ones. Consider 

(x, y) = T(r, 9) = (r cos#, r sin#). 

Note that T is not one-one since T(0, 0) = (0, 0) = T(0, n). (Indeed T(0, 9) = 
(0, 0) for all real numbers 9.) Note that vertical lines in the r0-plane given by 
r = a, where a is constant, are mapped to the points (x, y) = (a cos#, a sin#) 
on a circle of radius a. Horizontal rays {(r, 9) \ 9 = a, r > 0} are mapped to 
rays emanating from the origin. (See Figure 5.81.) It follows that the rectangle 
D* = [5, 1] x [0, it] in the r#-plane is mapped not to a parallelogram, but bent 



6 





G = a 


r = a 













Image of 
6 = a 




Image of 
r = a 



Figure 5.81 The images of lines in the r#-plane under the 
transformation T(r, 6) = (r cos 0, r sin 0). 



5.5 | Change of Variables 355 



D 




Figure 5.82 The image of the rectangle D* = [±, 1] x [0, n] 
under T(r, 9) = (r cos 6, r sin#). 




Figure 5.83 The image of B* = [\, 1] x [0, it] x [0, 1] 
under T(r, 6, z) = (r cos 6, r sin#, z). 



into a region D that is part of the annular region between circles of radii \ and 1 , 
as shown in Figure 5.82. 

Analogously, the transformation T: R 3 — > R 3 given by 

(x, y, z) = T(r, 6, z) = (r cos#, r sin^, z) 

bends the solid box B* = [j, 1] x [0, 7r] x [0, 1] into a horseshoe-shaped solid. 
(See Figure 5.83.) ♦ 



Change of Variables in Definite Integrals 

Now we see what effect a coordinate transformation can have on integrals and how 
to take advantage of such an effect. To begin, consider a case with which you are 
already familiar, namely, the method of substitution in single-variable integrals. 

EXAMPLE 8 Consider the definite integral / Q 2 2x cos(x 2 ) dx. To evaluate, one 
typically makes the substitution u = x 2 (so du = 2x dx). Doing so, we have 

sin 4. 



/ 2x cos(x 2 ) dx = I cosudu = sinu 
Jo Jo 



u=4 
u=0 



Let's dissect this example more carefully. First of all, the substitution u = x 2 
may be rewritten (restricting x to nonnegative values only) as x = *Ju. Then 
dx = du /(ly/u) and 

f 2 f 4 du f 4 

/ 2xcos(x 2 )dx= I 2yfu cos(a/m) 2 — — = / cosMJw = sin4. 
Jo Jo 2y/u Jo 

In other words, the substitution is such that the 2x = 2^fu factor in the integrand 
is canceled by the functional part of the differential dx = du/(2*Ju). Hence, a 
simple integral results. ♦ 



In general, the method of substitution works as follows: Given a (perhaps 
complicated) definite integral f A f(x)dx, make the substitutions = x(u), where 
x is of class C 1 . Thus, dx = x'(u)du. If A = x(a), B = x(b), and x'(u) 7^ 0 for 
u between a and b, then 

/• B nb 

I f(x)dx = J f(x(u))x'(u)du. (3) 

J A Ja 



Chapter 5 | Multiple Integration 



x 



x = x(u) 




It 



Figure 5.84 As Am = du — >• 0, 
Ax — > dx = x'(u) Am. Thus, the factor 
x'(u) measures how length in the w-direction 
relates to length in the x-direction. 



Note that it is possible to have a > b in (3) above. (This happens if x(u) is 
decreasing.) Although the m -integral in equation (3) may at first appear to be more 
complicated than the x -integral, Example 8 shows that in fact just the opposite 
can be true. 

Beyond the algebraic formalism of one-variable substitution in equation (3), 
it is worth noting that the term x'(u) represents the "infinitesimal length distortion 
factor" involved in the changing from measurement in u to measurement in x. 
(See Figure 5.84.) We next attempt to understand how these ideas may be adapted 
to the case of multiple integrals. 

The Change of Variables Theorem for Double Integrals 

Suppose we have a differentiable coordinate transformation from the wu-plane to 
the xy-plane. That is, 

T: R 2 -> R 2 , T(w, v) = (x(u, v), y(u, v)). 



DEFINITION 5.2 The Jacobian of the transformation T, denoted 


d(x, y) 




d(u, v) ' 




is the determinant of the derivative matrix DT(u, 


v). That is, 




dx dx 




d(x ' y) =detDT(u, u) = det 

d(u, v) 


du dv 
dy dy 
du dv 


dx dy dx dy 
du dv dv du 



The notation d(x, y)/d(u, v) for the Jacobian is a historical convenience. The 
Jacobian is not a partial derivative, but rather the determinant of the matrix of 
partial derivatives. It plays the role of an "infinitesimal area distortion factor" 
when changing variables in double integrals, as in the following key result: 



5.5 | Change of Variables 



y 



(f.f) 




(8,0) 

Figure 5.85 The triangular region 
D of Example 9. 



THEOREM 5.3 (Change of variables in double integrals) Let D and D* 
be elementary regions in (respectively) the xy-plane and the Mu-plane. Suppose 
T:R 2 — > R 2 is a coordinate transformation of class C 1 that maps D* onto D 
in a one-one fashion. If /: D — > R is any integrable function and we use the 
transformation T to make the substitution x = x(u, v), y = y(u, v), then 



jj f(x,y)dxdy = j j f(x(u,v),y(u,v)) 



dp, y) 
d(u, v) 



du dv. 



EXAMPLE 9 We use Theorem 5.3 to calculate the integral 



cos(x + 2y) sin(x — y) dx dy 



over the triangular region D bounded by the lines y = 0, y = x,andx + 2y = 8 as 
shown in Figure 5.85. It is possible to evaluate this integral by using the relatively 
straightforward methods of §5.2. However, this would prove to be cumbersome, 
so, instead we find a suitable transformation of variables, motivated in this case by 
the nature of the integrand. In particular, we let u = x + 2y, v = x — y. Solving 
for x and y, we obtain 



u + 2v 
x = and 



y 



Therefore, 



d(x, y) 
d(u, v) 



det 



X u Xy 

y u y v 



det 



I 2 

3 3 

1 _ 1 

3 3 



Considering the coordinate transformation as a mapping T(w , v) = (x, y) of 
the plane, we need to identify a region D* that T maps in a one-one fashion 
onto D. To do this, essentially all we need do is to consider the boundaries of D : 



y = x 
x + 2y = 8 

y = 0 



x - y = 0 
u = 8; 

u — v 
= 0 



v = 0; 



v = u. 



Hence, one can see that T transforms the region D* shown in Figure 5.86 onto 
D. Therefore, applying Theorem 5.3, 




Figure 5.86 The effect of the transformation T of Example 9. 



Chapter 5 | Multiple Integration 



j j cos(x + 2 v) sin(x — y)dx dy = j j cos u sin v 



B(x, y) 



du dv 



d(u, v) 

= jj cosw sinv | — 1| dudv 

= J J j cos u sin v dv du 
Jo Jo 

,8 

= / | cos u ( — cos v)| JJZq du 
Jo 

= | / cosw(— cosm + 1) du 
Jo 

, f* 

= 3/ (cos u — cos u) du 
Jo 

sinwlg — j j(1+cos2m)Jm 
sin 8 — [ju + \ sin2M)|pJ 



= \ [sin8 - 4 



sin 



16] 



There is another, faster way to calculate the Jacobian, namely, to calculate 
d(u, v)/d(x, y) directly from the variable transformation, and then to take reci- 
procals. That is, from the equations u = x + 2y, v = x — y, we have 



d(u, v) 
d(x, y) 



det 



I'v 



Uy 
Vy 



det 



-3. 



Consequently, 3(x, y)/d(u, v) = — |, which checks with our previous result. 
This method works because if T(h, v) = (x, v), then, under the assumptions of 
Theorem 5.3, (u,v) = T _1 (jc, y). It follows from the chain rule that 



DT-\x,y)= [DT(u,v)]-\ 

(That is, DT 1 is the inverse matrix of DT. See Exercises 30-38 in § 1 .6 for more 
about inverse matrices.) Hence, 



^=det[DT- 1 ]=det[(DTr 1 ] 

d(u, v) 



detOT 



EXAMPLE 10 We use Theorem 5.3 to evaluate / f D (x 2 - y 2 ) e xy dx dy, where 
D is the region in the first quadrant bounded by the hyperbolas xy = 1, xy = 4 
and the lines y = x, y = x + 2. (See Figure 5.87.) 



5.5 | Change of Variables 359 



y y=x+2 




Figure 5.87 The region D of Example 10. 



D* 



Figure 5.88 The region D* 
corresponding to the region D of 
Example 10. 



Both the integrand and the region present complications for evaluation. There 
would seem to be two natural choices for ways to transform the variables. One 
would be 

u = x 2 — y 2 and v = xy, 

motivated by the nature of the integrand. However, the region D of integration 
will not be easy to describe in terms of this particular choice of wu-coordinates. 
Another possible transformation of variables, motivated instead by the shape of 
D, is 

u = xy and v = y — x . 

Now this change of variables would not seem to help much with the integrand, 
but, as we shall see, it turns out to be just what we need. 

First note that the boundary hyperbolas xy = 1 and xy = 4 correspond re- 
spectively, to the lines u = 1 and u = 4; the lines y = x and y = x + 2 correspond 
to v = 0 and v = 2. Thus, the region D* in the Mu-plane that corresponds to D 
(see Figure 5.88) is 

D* = {(u, v) | 1 < u < 4, 0 < v < 2}. 
Next, we calculate that the Jacobian of the variable transformation is 



9(w, u) 



= det 



y x 
-1 1 



= x + y. 



d(x, y) 

Hence, the Jacobian we require in order to use Theorem 5.3 is 

d(x, y) = 1 
9(w, v) x + y 

Moreover, since we will be working in the first quadrant (where x and y are both 
positive), |3(jc, y)/9(w, v)\ = l/(x + y). 



Chapter 5 | Multiple Integration 



At last we are ready to compute: 



jji* 2 - y 2 ) e * y dx d y = jj (x 2 - y 2 ) e * y (v^) du dv 
= f 2 C (*-?) (* + y) 

Jo J i x 



+ y 

du dv 



e xy dudv 



-u(e 4 - e l )dv = -y 0 - e ) 



2 

k' 



2(e - A 



Note that the insertion of the Jacobian in the integrand caused precisely the can- 
celation needed to make the evaluation straightforward. We cannot always expect 
this to happen, but the lesson here is to be willing to carry through calculations 
that may not at first appear to be so easy. ♦ 

EXAMPLE 11 (Double integrals in polar coordinates) In Example 9, a coor- 
dinate transformation was chosen primarily to simplify the integrand of the double 
integral. In this example we change variables by using a coordinate system better 
suited to the geometry of the region of integration. 

For example, suppose that the region D is a disk of radius a : 

D={(x,y)\x 2 + y 2 < a} 

= |( x > y) I — V a 2 — x 2 < y < \l a 1 — x 2 , —a < x < a J . 

Then, to integrate any (integrable) function / over D in Cartesian coordinates, 
one would write 

/ / f(x,y)dxdy= I I f(x,y)dydx. 

J Jd J -a J-Ja^l? 

Even if it is easy initially to find a partial antiderivative of the integrand, the limits 
in the preceding double integral may complicate matters considerably. This is be- 
cause the disk is described rather awkwardly by Cartesian coordinates. We know, 
however, that it has a much more convenient description in polar coordinates as 

{(r, 0) | 0 <r <a,0 <6 < lit). 

This suggests that we make the change of variables 

(x, y) = T(r, 0) = (r cos#, r sin#), 

which is shown in Figure 5.89. (Note that T maps all points of the form (0, 6) to 
the origin in the xy-plane and thus, cannot map D* in a one-one fashion onto D. 
Nonetheless, the points of D* on which T fails to be one-one fill out a portion 
of a line — a one-dimensional locus — and it turns out that it will not affect the 
double integral transformation.) The Jacobian for this change of variables is 



^ = det 
d(r, 6) 



cos 9 —rsinO 
sin 0 r cos 9 



r cos 2 9 + r sin 2 9 = r. 



2k 








D* 






a 



5.5 | Change of Variables 

y 



361 





-~^x 2 + y 2 


A 






) 







Figure 5.89 T maps the (nonclosed) rectangle D* to 
the disk D of radius a. 







< 













Figure 5.90 The disk of 
radius a centered at the 
origin. 



(0,2) 



x 2 + y 2 = 4 




Figure 5.91 The region 
D of Example 13. 



(Note that r > 0 on D, so \r\ — r.) Thus, using Theorem 5.3, the double integral 
can be evaluated by using polar coordinates as follows: 



y) dx dy 



/a p \J a 2 —x 2 
/ f(x,y)dydx 
-a J—fa^ 1 



p2n pa 
JO JO 



f(r cos#, r sin9)r dr d9. 



It is evident that the limits of integration of the r0-integrals are substantially 
simpler than those in the xy-integral. Of course, the substitution in the integrand 
may result in a more complicated expression, but in many situations this will not be 
the case. Polar coordinate transformations will prove to be especially convenient 
when dealing with regions whose boundaries are parts of circles. ♦ 

EXAMPLE 12 To see polar coordinates "in action," we calculate the area of a 
circle, using double integrals. Once more, let D be the disk of radius a, centered 
at the origin as in Figure 5.90. Then we have 



Area 



r- r- f a f V« 2 — X 2 pin pa 

ldA= / dydx = 

J JD J -a J-Ja 2 -x 2 JO Jo 



r dr d9, 



following the discussion in Example 1 1 . The last iterated integral is readily eval- 
uated as 



j jrdrd9 = j (jr 2 \tyd9 = \a 2 dd = \a 2 (2Tt - 0) = ita 2 , 

which indeed agrees with what we already know. If you feel so inclined 
compare this calculation with the evaluation of the iterated integral in Cartesian 
coordinates. No doubt you'll agree that the use of polar coordinates offers clear 
advantages. ♦ 



EXAMPLE 13 We evaluate the double integral ff D x 2 + y 2 + ldx dy, 

where D is the quarter disk shown in Figure 5.91, using polar coordinates. The 
region D of integration is given in Cartesian coordinates by 



so that 



D = {(x, y) | 0 < y < y^-x 2 , 0 < x < 2}, 



f f yjx 2 + y 2 + \dxdy = f f y 1 x 2 + y 2 + \dydx 
J Jd Jo Jo 



Chapter 5 | Multiple Integration 



This iterated integral is extremely difficult to evaluate. However, D corresponds 
to the polar region 

D* = {(r, 9)\0<r <2, 0 < 6> < jt/2}. 
Therefore, using Theorem 5.3, we have 

J J Vjc 2 + y 2 + 1 dx dy = ff^> 



r 2 cos 2 9 + r 2 sin 2 9 + 1 • r dr d6 



r 2 +\rdrd9 



= r ' n, 

Jo Jo 

= / |(r 2 +l) 3 / 2 |_ 0 ^ 
Jo 

-L 



i(5 3/2 - \)dd 
- !)■ 



Sketch of a proof of Theorem 5.3 Let (uq, vq) be any point in D* and let 
Au = u — uo, Av = v — vo- The coordinate transformation T maps the rectangle 
R* inside D* (shown in Figure 5.92) onto the region R inside D in the xy-plane. 
(In general, R will not be a rectangle.) Since T is of class C 1 , the differentiability 
of T (see Definition 3.8 of Chapter 2) implies that the linear approximation 



h(u, v) = T(w 0 , v 0 ) + DT(m 0 , Wo) 
= T(u 0 , vq) + DT(u 0 , u 0 ) 



U — Mo 

v - V 0 



Au 
Av 




is a good approximation to T near the point («o , fo)- In particular, h takes the rect- 
angle R* onto some parallelogram P that approximates R as shown in Figure 5.93. 
We compare the area of R* to that of P. 

From Figure 5.93, we see that the rectangle R* is spanned by 

0 



Am 
0 



and b 



5.5 | Change of Variables 




P = h(R*) 



DT(u 0 , v 0 )b 




R = T(R') 



T(w 0 , 

h( Mo , v 0 ) ^DT(u 0 , u 0 )a 



Figure 5.93 The linear approximation h takes the rectangle R* onto a parallelogram P 
that approximates R = T(R*). 



and the parallelogram P is spanned by the vectors c = DT(mo, t>o)a and d = 
DT(wo, wo)b. Hence, 



Area of 7?* = ||a x b|| = Am Av, 
and thus, by Proposition 5.1, 

d(x, y) 



Area of P — lie x dll 



detDT(w 0 , Vo)\Au Av 



d(u, v) 



(u 0 , v 0 ) 



Au Av. 



This result gives us some idea how the Jacobian factor arises. 

To complete the sketch of the proof, we need a partitioning argument. Partition 
D* by subrectangles Rf, . Then we obtain a corresponding partition of D into (not 
necessarily rectangular) subregions Rjj = T(7?* ). Let A Ajj denote the area of Rjj . 
Let Cij denote the lower left corner of /?*■ and let d i; = T(c, 7 ). (See Figure 5.94.) 
Then, since / is integrable on D, 

/ / f(x,y)dxdy= lim V/(d, 7 )AA i7 . 

J J D all Rij-tO *7-f 



From the remarks in the preceding paragraph, we know that 

d(x,y) 



AAjj «a area of parallelogram h(/?* ) = 



d(u, v) 



(c i7 ) 



Aui Avj. 




Figure 5.94 A partition of D* gives rise to a partition of D. 



364 Chapter 5 | Multiple Integration 




Figure 5.95 The "area element" 
d A in rectangular coordinates is 
dx dy. 



D' 



x = r cos 6 
y = r sin 0 




Figure 5.96 The polar-rectangular transformation takes rectangles in the 
r#-plane to wedges of disks in the Jty-plane. 



Arclength = rAd 




Figure 5.97 An infinitesimal 
polar wedge. 



Taking limits as all the Rjj tend to zero (i.e., as Am, and Avj approach zero), we 
find that 



[ [ f(x,y)dxdy = hm Tf (T(c, 7 )) 



d(x, y) 



d(u, v) 



(«y) 



All: AVj 



If. 



f(x(u, v), y(u, u)) 



d(x, y) 



d(u, v) 



du dv, 



as was to be shown. 



Consider again the polar-rectangular coordinate transformation. When we use 
Cartesian (rectangular) coordinates to calculate a double integral over a region 
D in the plane, then we are subdividing D into "infinitesimal" rectangles having 
"area" equal to dx dy. (See Figure 5.95.) On the other hand when we use polar 
coordinates to describe this same region, we are subdividing D into infinitesimal 
pieces of disks instead. (See Figure 5.96.) These disk wedges arise from trans- 
formed rectangles in the r#-plane. One such infinitesimal wedge in the xy-plane 
is suggested by Figure 5.97. When AG and Ar are very small, the shape is nearly 
rectangular with approximate area (r AO) Ar. Thus, in the limit, we frequently say 



d A = dx dy (Cartesian area element) 
= r dr d6 (polar area element). 



Change of Variables in Triple Integrals 

It is not difficult to adapt the previous reasoning to the case of triple integrals. We 
omit the details, stating only the main results instead. 



DEFINITION 5.4 Let T: R 3 -> R 3 be a differentiable coordinate transfor- 
mation 

T(m, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w)) 



5.5 | Change of Variables 



from www-space to xvz-space. The Jacobian of T, denoted 

90, y, z) 



d(u, v, w)' 



is det(DT(w, v, w)). That is, 



d(x, y, z) 



d(u, v, w) 



= det 



3x 


dx 


dx 


du 


dv 


dw 


3? 


dy 


dy_ 


du 


dv 


dw 


dz 


dz 


dz 


_ du 


dv 


dw _ 



In general, given any differentiate coordinate transformation T: R" 
the Jacobian is just the determinant of the derivative matrix: 



R" 



d(x u ...,x„) 



d{u\ 



detDT(M! 





3xi 


3xi 


3xi 




du\ 


du 2 


du„ 




dxi 


3x 2 


dx 2 


det 


du\ 


du 2 


du n 




dx n 


dx„ 


dx n 




du\ 


dU2 


du n 



THEOREM 5.5 (Change of variables in triple integrals) Let W and W* be 

elementary regions in (respectively) xvz-space and wuw-space, and let T: R 3 -> 
R 3 be a coordinate transformation of class C 1 that maps W* onto W in a one-one 
fashion. If /: W — > R is integrable and we use the transformation T to make the 
substitution x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), then 



/ / / f{x,y,z)dxdydz 
J J Jw 



I I Iw* 



f(x(u, v, w), y(u, v, w), z(u, v, w)) 



d(x, y, z) 
d(u, v, w) 



du dv dw. 



(See Figure 5.98.) 



In the integral formula of the change of variables theorem (Theorem 5.5), the 
Jacobian represents the "volume distortion factor" that occurs when the three- 
dimensional region W is subdivided into pieces that are transformed boxes in 
Muw-space. (See Figure 5.99.) In other words, the differential volume elements 



Chapter 5 | Multiple Integration 




Figure 5.98 A three-dimensional transformation T that takes the solid 
region W* in uvw-space to the region W in xyz-space. 



(i.e., "infinitesimal" pieces of volumes) in xyz- and Muw-coordinates are related 
by the formula 



dV = dx dy dz = 



d(x, y, z) 
d(u, v, w) 



dudvdw. 




Figure 5.99 The volume of the "infinitesimal box" in 
uvw-space is du dv dw. The image of this box under T 
has volume \d(x, y, z)/d(u, v, w)\ du dv dw. 



EXAMPLE 14 (Triple integrals in cylindrical coordinates) When integrat- 
ing over solid objects possessing an axis of rotational symmetry, cylindrical 
coordinates can be especially helpful. The cylindrical-rectangular coordinate 
transformation 

' x = r cos 9 
y = r sin 6 

z = z 



has Jacobian 



d(x, y, z) 



= det 



cost' 
sinO 
0 



-r sind 
r cos 9 
0 



= r cos 9 + r sin 9 = r. 



d(r, 9, z) 

Hence, the formula in Theorem 5.5 becomes 

f(r cos6», r sm.9, z)r dr d9 dz. 



Ill fix, y, z)dxdydz = 
J J Jw J J JV\ 



5.5 | Change of Variables 367 

In particular, we see that the volume element in cylindrical coordinates is 

dV = rdrd6dz. 

(Recall that the cylindrical coordinate r is usually taken to be nonnegative. Given 
this convention, we may omit the absolute value sign in the change of variables 
formula.) The geometry behind this volume element is quite plausible: A "dif- 
ferential box" in r#z-space is transformed to a portion of a solid cylinder that is 
nearly a box itself. (See Figure 5. 100.) ♦ 




Figure 5.1 00 A "differential box" in rOz-space is mapped to a portion of a 
solid cylinder in xyz-space by the cylindrical-rectangular transformation. 



EXAMPLE 15 To calculate the volume of a cone of height h and radius a, we 
may use Cartesian coordinates, in which case the cone is the solid W bounded by 
the surface az = h^/x 2 + y 2 and the plane z = h, as shown in Figure 5.101. The 
volume can be found by calculating the iterated triple integral 



/a p ~J a 1 — x 2 p 
-a J—Ja^x 2 J% 



dz dy dx. 



We will forgo the details of the evaluation, noting only that trigonometric substi- 
tutions are necessary and that they make the resulting computation quite tedious. 



az 





Shadow in xy-plane 



Figure 5.101 The solid cone W of Example 15. 



In contrast, since the cone has an axis of rotational symmetry, the use of 
cylindrical coordinates should afford us substantially less involved calculations. 



Chapter 5 | Multiple Integration 




Figure 5.102 The cone of 
Example 15 described in 
cylindrical coordinates. 



Hence, we consider the cone again. (See Figure 5.102.) Note that 

h 



W = 



(r, 9, z) 



r < z <h, 0 < r <a, 0 < 9 < lit 



Thus, the volume is given by 



n n n p 7,71 no nh 

dV = / / rdzdrdO. 

J J Jw J0 J0 J±r 



(Note the order of integration that we chose.) The evaluation of this iterated 
integral is exceedingly straightforward; we have 



/ / / rdzdrd0= / / 

Jo Jo 3\r Jo Jo 

-L 

-re- 



2tt P a / h 

r\h- -r\drd6 



h j h 

—r 1 

2 3a 



dO 



a x d6 



= 2tc I —a' 
6 



= —a h, 



which agrees with what we already know. 



EXAMPLE 16 (Triple integrals in spherical coordinates) Ifa solid object has 
a center of symmetry, then spherical coordinates can make integration over such 
an object more convenient. The spherical-rectangular coordinate transformation 

x = p sin <p cos 6 
y = p sin cp sin 9 
z = p cos cp 



has Jacobian 

d(x, y, z) 

B(p,<p,e) 



= det 



simp cost 
sin<p sin£ 
coscp 



p coscp cos( 
p cos cp sin ( 
— p sin^ 



-p sin<p sin# 
p sin <p cos 0 
0 



Using cofactor expansion about the last row, this determinant is equal to 
cos (p (p 2 cos 2 9 sin <p cos cp + p 2 sin 2 9 sin cp cos <p) 



+ p sin cp (p cos 2 9 sin 2 (p + p sin 2 1 



sin 



= p 2 cos <p(sin (p cos cp) + p 2 sin 3 <p 
= p 2 sin cp (cos 2 cp + sin 2 (p) 
= p 2 sincp. 

(Under the restriction that 0 < <p < jt, sin^) will always be nonnegative. Hence, 
the Jacobian will also be nonnegative.) Therefore, the volume element in spherical 



5.5 | Change of Variables 



de 



dq> 



/dp 




rdd = psin(p dd 



Figure 5.1 03 A differential box in p<p#-space is mapped to a portion of a solid ball 
in xyz-space by the spherical-rectangular transformation. 



coordinates is 



dV = p sin<p dp d<p dO, 



and the change of variables formula in Theorem 5.5 becomes 

/ / / /(*> y> z)dxdydz 
J J Jw 



-III 



f(x(p, ip, 0), y(p, <p, 0), z(p, (p, 0))p sirup dp dip dO. 



The volume element in spherical coordinates makes sense geometrically, because 
a differential box in p(p0 -space is transformed to a portion of a solid ball that is 
approximated by a box having volume p 2 sin q> dp d(p dO. (See Figure 5. 103.) ♦ 

EXAMPLE 17 The volume of a ball is easy to calculate in spherical coordi- 
nates. A solid ball of radius a may be described as 

B = {(p, (p,9)\0<p<a,0<(p<7T,0<e< 2jt}. 

(See Figure 5 . 1 04.) Hence, we may compute the volume by using the triple integral 



fff dV = f ! ' r r 

J J Jb Jo Jo Jo 



p sirup dp d(p d9 



, . 3 
^0 ^0 



sin<p dcp dO 



- (-coscp\*)d0 = - (-(-l) + l)d0 
j Jo j Jo 



2a 



3 flit 



dO 



as expected. 



EXAMPLE 18 We return to the example of the cone of radius a and height h 
and this time, use spherical coordinates to calculate its volume. First, note two 
things : (i) that the cone 's lateral surface has the equation (p = tan~ 1 (a / h) in spher- 
ical coordinates and (ii) that the planar top having Cartesian equation z = h has 
spherical equation p cos cp = h or, equivalently, p = h sec (p. (See Figure 5. 105.) 



Chapter 5 | Multiple Integration 



For fixed values of the spherical angles tp and 0, the values of p that give 
points inside the cone vary from 0 to h sec (p. Any points inside the cone must 
have spherical angle <p between 0 and tan~'(a/ h). Finally, by symmetry, 0 can 
assume any value between 0 and 2n , Hence, the cone may be described as the set 



0 < p < h sec<p, 0 < (p < tan 1 '-, 0 < 0 < 2n \ 

h 



Therefore, we calculate the volume as 

"2jt /»tan (a/H) phsecy 



pin /'tan \ajn) p 

Jo Jo Jo 

=/7 

^0 ^0 



p sin <p dp d<p d0 
/h) (h sec<p) 3 



sin<pd(p dO 



^3 r2n ntarT l (a/h) 



h i pm pi 

= — / / sec J cp siri(p d(p dO 

3 Jo Jo 

^3 pin ptsar l (alh) 

= — / / tmcp sec 2 cp dtp dO. 

3 Jo Jo 

Now, let u = tany> so du = sec 2 (p dep. Then the last integral becomes 

h 3 f 2n [ a,h h 3 f 2n 1 /a\ 2 h 3 a 2 f 2lt 

— / ududO = — - - ) dO = — - / dO 
3 Jo Jo 3 Jo 2 \hj 6h* Jo 



3 

a 2 h 



-{lit) 



Tt 



-a h, 



as expected. 



The use of spherical coordinates in Example 18 is not the most appropri- 
ate. We merely include the example so that you can develop some facility with 
"thinking spherically." Further practice can be obtained by considering some of 
the applications in the next section as well as, of course, some of the exercises. 

Summary: Change of Variables Formulas 



Change of variables in double integrals: 

jj f(x,y)dxdy = j j f(x(u,v),y(u,v 



)) 



d(x, y) 



d(u, v) 



du dv 



Area elements: 



dA = dx dy 
= rdrd0 

djpc, y) 
d(u, v) 



(Cartesian) 
(polar) 

dudv (general) 



5.5 I Exercises 371 



Change of variables in triple integrals: 

/ / / /(*> J> z)dxdydz 
J J Jw 



=fll 



f(x(u, v, w), y(u, v, w), z(u, v, w)) 



d(x, y, z) 



d(u, v, w) 



du dvdw 



Volume elements: 

dV = dxdydz (Cartesian) 
= r dr d6 dz (cylindrical) 
= p 2 sin (p dp d<p d9 (spherical) 
3(x, y, z) 



d(u, v, w) 



dudvdw (general) 



5.5 Exercises 



1. LetT(M, v) = (3m, -v). 
(a) Write T(u, v) as A 



for a suitable matrix A. 



(b) Describe the image D = T(D*), where D* is the 
unit square [0, 1] x [0, 1]. 



2. (a) Let 



T(m, v) 



u — v u + v 



V5 ' V2 



How does T transform the unit square D* 
[0, 1] x [0, 1]? 

(b) Now suppose 



T(k, v) 



U + V u — V 



V2 ' V2 
Describe how T transforms D*. 



3. If 



T(m, v) = 



2 3 
-1 1 



and D* is the parallelogram whose vertices are (0, 0), 
(1, 3), (-1, 2), and (0, 5), determine D = T(D*). 

4. If D* is the parallelogram whose vertices are (0, 0), 
(—1,3), (1, 2), and (0, 5) and D is the parallelogram 
whose vertices are (0, 0), (3, 2), (1, —1), and (4, 1), 
find a transformation T such that T(Z)*) = D. 

5. If T(m, v, w) = (3m — v, u — v + 2w, 5m + 3d — w), 
describe how T transforms the unit cube W* = 
[0, 1] x [0, 1] x [0, 1]. 



6. Suppose T(m, d) = (u, uv). Explain (perhaps by us- 
ing pictures) how T transforms the unit square D* = 
[0, 1] x [0, 1]. Is T one-one on D*l 

7. Let T: R 3 — >■ R 3 be the transformation given by 

T(p, <p, 6) = (p sin <p cos 6, p sin<p sin 9, p cos^). 

(a) Determine D = T(D*), where D* = [0, 1] x 
[O.tt] x [0,2jt]. 

(b) Determine D = T(D*), where D* = [0, 1] x 
[0, 7T/2] x [0, tt/2]. 

(c) Determine D = T(D*), where D* = [1/2, 1] x 
[0, tt/2] x [0, TT/2]. 

8. This problem concerns the iterated integral 



J0 Jy/2 



(2x — y)dx dy. 



(a) Evaluate this integral and sketch the region D of 
integration in the .v v-plane. 

(b) Let u = 2x — y and v = y. Find the region D* in 
the MD-plane that corresponds to D. 

(c) Use the change of variables theorem (Theorem 5.3) 
to evaluate the integral by using the substitution 

u = 2x — y, v = y. 



9. Evaluate the integral 

r2 |-(x/2)+l 

x/2 

by making the substitution u = x, v = 2y — x. 



ff 

Jo Jxl 



x 5 (2y -x)e (2y ~ x)2 dydx 



Chapter 5 | Multiple Integration 



10. Determine the value of 



x + y 



2y 



dA, 



where D is the region in R 2 enclosed by the lines 
y = x/2, y = 0, and x + y = 1 . 

1 1 . Evaluate f f D (2x + y) 2 e x ~ y dA, where D is the region 
enclosed by 2x +y = 1, 2x + y = 4, x — y = — 1, 
and x — y = 1 . 



12. Evaluate 



SL 



(2x + y-3) 2 

dx dy, 

D (2y -x + 6) 2 



where D is the square with vertices (0, 0), (2, 1), 
(3, -1), and (1, -2). (Hint: First sketch D and find 
the equations of its sides.) 

In Exercises 13—1 7, transform the given integral in Cartesian 
coordinates to one in polar coordinates and evaluate the polar 
integral. 



3 dy dx 



dy dx 



Jo Jo 

15. J j (x 2 + y 2 ) 3/2 dA, where D is the disk* 2 + v 2 <9 

/a r^fa 1 
-a JO 

f'f 

Jo Jo 



dx dy 



dy dx 
Jx 2 + y 2 



18. Evaluate 



11 



d ^4" 



dA, 



where D is the disk of radius 1 with center at (0, 1). 
(Be careful when you describe D.) 

19. Let D be the region between the square with vertices 
(1, 1),(-1, 1),(-1, -1),(1, -1) and the unit disk cen- 



tered at the origin. Evaluate 



y 2 dA. 



20. Find the total area enclosed inside the rose r = sin 26 . 
(Hint: Sketch the curve and find the area inside a single 
leaf.) 

21. Let n be a positive integer, and let a be a posi- 
tive constant. Calculate the total area inside the rose 
r = a cosn8 and show that the value depends only on 
a and whether n is even or odd. 



22. Find the area of the region inside both of the circles 
r = 2a cos 6 and r = 2a sin 9, where a is a positive 
constant. 

23. Find the area of the region inside the cardioid r = 
1 — cos 8 and outside the circle r = 1 . 

24. Find the area of the region bounded by the positive 
x-axis and the spiral r = 39, 0 < 6 < 2n. 



25. Evaluate 



cos(x 2 + y 2 )dA, 



where D is the shaded region in Figure 5.106. 



Arc of a circle 
of radius 1 
(centered at 
origin) 




Figure 5.106 The region D of 
Exercise 25. 



26. Evaluate j j sin (x 2 + y 2 )dA, where D is the region 

in the first quadrant bounded by the coordinate axes 
and the circles x 2 + y 2 = 1 and x 2 + y 2 = 9. 



27. Use polar coordinates to evaluate 



where D is the unit square [0, 1] x [0, 1]. 

*3 /.V9-JC 2 z-3 



dA, 



/) fV-r pi 
I j 
-3 J -^9=3? J \ s /x 2 +y 1 

ina 

/l r-s/\-y 2 r4 
-1 J-Jl-v 2 JO 



using cylindrical coordinates. 



Jx 2 + y 



dz dy dx by 



29. Evaluate 



using cylindrical coordinates. 
30. Evaluate 

dV 



■ dz dx dy by 



///. 



Ib y/x 2 + y 2 + z 2 + 3 ' 
where B is the ball of radius 2 centered at the origin. 
31. Determine 

(x 2 + y 2 +2z 2 )dV, 



where W is the solid cylinder defined by the inequali- 
ties x 2 + y 2 < 4, - 1 < z < 2. 



5.6 | Applications of Integration 



32. Determine the value of 



SSL 



dV ', where 



>w jx 2 + y 2 

W is the solid region bounded by the plane z = 12 and 
the paraboloid z = 2x 2 + 2y 2 — 6. 

33. Find the volume of the region W bounded on top by 
z = y/a 2 — x 2 — y 2 , on the bottom by the xy-plane, 
and on the sides by the cylinder x 2 + y 2 = b 2 , where 

0 < b < a. 

In Exercises 34 and 35, determine the values of the given in- 
tegrals, where W is the region bounded by the two spheres 
x 2 + y 2 + z 2 = a 2 and x 2 + y 2 + z 2 = b 2 , for 0 < a < b. 



«• SSL 



dV 



35 



w jx 2 + y 2 + z 2 

■III* 



x 1 + y L + z z e x 



36. Let W denote the solid region in the first octant be- 
tween the spheres x 2 + y 2 + z 2 = a 2 and x 2 + y 2 + 
z 2 = b 2 , where 0 < a < b. Determine the value of 

fff w (x + y + z)dv. 

37. Determine the value of fff w z 2 dV, where W is the 
solid region lying above the cone z = ^3x 2 + 3y 2 and 
inside the sphere x 2 + y 2 + z 2 = 6z. 



38. Determine 



where W 



(l + Jx 2 + y 2 ) dV, 



(x,y,z)\ y/x 2 + y 2 <z/2<3\. 



39. Find the volume of the region W that represents the 
intersection of the solid cylinder x + y 2 < 1 and the 
solid ellipsoid 2(x 2 + y 2 ) + z 2 < 10. 

40. Find the volume of the solid W that is bounded by 
the paraboloid z = 9 — x 2 — y 2 , the xy-plane, and the 
cylinder x 2 + y 2 = 4. 



41. Find 



SSL 



(2 + x 2 + y 2 )dV, 



where W is the region inside the spheres 2 + y 2 + z 2 = 
25 and above the plane z = 3. 

42. Find the volume of the intersection of the three solid 
cylinders 

x 2 + y 2 <a 2 , x 2 + z 2 <a 2 , and y 2 +z 2 <a 2 . 

(Hint: First draw a careful sketch, then note that, by 
symmetry, it suffices to calculate the volume of a por- 
tion of the intersection. ) 



Day 


°F 


Monday 


65 


Tuesday 


63 


Wednesday 


52 


Thursday 


51 


Friday 


45 


Saturday 


43 


Sunday 


47 



5.6 Applications of Integration 

In this section, we explore a variety of settings where double and triple integrals 
arise naturally. 

Average Value of a Function 

Suppose temperatures (shown in the adjacent table) are recorded in Oberlin, Ohio, 
during a particular week. From these data, we calculate the average (or mean) 
temperature: 

65 + 63 + 52 + 51 +45 + 43 + 47 



Average temperature 



52.3°F. 



Of course, this calculation only represents an approximation of the true average 
value, since the temperature will vary during each day. To determine the true 
average temperature, we need to know the temperature as a function of time for 
all instants of time during that one-week period; that is, we consider 

Temperature = T(x), x = elapsed time (in days), for 0 < x < 7. 

Then a more accurate determination of the average temperature is as an integral: 



Average temperature 



= 1 fn* 

Jo 



) dx. 



(1) 



Since an integral is nothing more than the limit of a sum, it's not hard to see that 
the preceding formula is a generalization of the original discrete sum calculation 
to the continuous case. (See Figure 5.107.) 



374 Chapter 5 | Multiple Integration 



80 T 




10-- 



0 1 2 3 4 5 6 
Days 

Figure 5.1 07 A continuous temperature function T(x) 
over the interval [0, 7]. The average temperature for the 
week is ^ / Q 7 T(x)dx. 



Note that 



7 = / dx = length of time interval. 
Jo 

Hence, we may rewrite formula (1) as 

Jq T(x)dx 



Average temperature = 



fo dx 



This observation leads us to make the following definitions concerning average 
values of functions. 



DEFINITION 6.1 (a) Let /: [a, b]^Rbe an integrable function of one 
variable. The average (mean) value of / on [a, b] is 



i" ft W - faf^ dX _ £ /W* 

U\«*-u ,1 J Wdx- "length of interval [a, i] - 

J a 



-f 

-a J a 



(b) Let /: D c R 2 -> R be an integrable function of two variables. The 
average value of / on D is 

m _ If D fdA _ ff D fdA 

ff D dA ~ areaof/3' 

(c) Let /: W c R 3 — > R be an integrable function of three variables. The 
average value of / on W is 

m IffwfdV ffUfdV 

L/Javg 



Jff w dV volume of W 



EXAMPLE 1 Suppose that the "temperature function" for Oberlin during a 
week in April is 

T(r\ — JIlv 7 - Mr 6 -I- 1127 r 5 _ 2393 4 , 66821 3 _ 45781 2 , 12581 , 

J W — 5040 A 180"* 180 A 72 * " 1 " 720 "* 360 A " 1 " 210 * " 1 " UJ ' 



5.6 | Applications of Integration 



(0,1) 




(2,0) 



Figure 5.108 The triangular 
metal plate of Example 2. 



where 0 < x < 7. Then the mean temperature for that week would be 



ITJavg — 7 1 0 / (5040* 7 
JO 



107 6 1 1127 5 
180 180 



2393 4 
72 X 



_i_ 66821 3 
720 X 

1 ( 113 „8 



45781 „2 , 12581 



360 



210 



x + 65) c/x 



40320 

, 66821 4 
2880 A 

888709 
17280 



107 7 1 1127 6 
1260 1080 



2393 5 
360 



^ + ^ + 65x)\l 



1080 

51.43°F 



EXAMPLE 2 Suppose that the thickness of the triangular metal plate, shown 
in Figure 5.108, varies as f(x, y) = xy + 1, where (x, y) are the coordinates of 
a point in the plate. The average thickness of the plate is, therefore, 

Jo fo~ 2y ( x y + l )dxdy 



Average thickness = 



fofo 2y dxd y 



Note that 



1 r2-2y 



II 

JO JO 



dx dy = area of triangular plate = ~(2 • 1) = 1 



from elementary geometry. Hence, the average thickness is 



fo fo ( 



xy+l)dxd\ I i\ 7 \|-*=2-2v , 

1 = / (2* y+ x )\ x =o d y 

Jo 

= f (\{2-2yf y + (2-2y))dy 
Jo 

= 2j\y 3 -2y 2 +l)dy = 2^- § y 3 + y) 



EXAMPLE 3 (See also Example 6 of §5.4.) Suppose the temperature inside the 
capsule bounded by the paraboloids z = 9 — x 2 — y 2 and z = 3x 2 + 3y 2 — 16 
varies from point to point as 

T(x,y,z) = z(x 2 + y 2 ). 

We calculate the mean temperature of the capsule. 
From Definition 6.1, 



[-^lavg — 



WwTdV 
IffwdV ' 



The particular iterated integrals we can use for the computation are then 

pS/2 ^25/4-x 2 ^-x 2 -y 2 

/ / / z(x 2 + y 2 )dzdydx 

J -5/2 J-^/25/4-A- 2 J3x 2 +3y 2 -l6 

[-^]avg = 



/5/2 p y/25/4-x 2 p9-x 2 -y 2 
I I dzdy dx 

-5/2 J-^/25/4-x 2 Av 2 +3j- 2 -16 



Chapter 5 | Multiple Integration 



Unfortunately, the calculations involved in evaluating these integrals are rather 
tedious. 

On the other hand, since the capsule has an axis of rotational symmetry, 
cylindrical coordinates can be used to simplify the computations. Note that the 
boundary paraboloids have cylindrical equations of z = 9 — r 2 and z = 3r 2 — 16 
and that the shadow of the capsule in the z = 0 plane can be described in polar 
coordinates as 

{(r,0) | 0 < r < §,0 < 9 < 2n) . 
(See Figures 5.109 and 5.1 10.) 




{(r,6)\0<r<j,Q<e<2n} 

y 



z = 3x 2 + 3.y 2 -16 
or 

z = 3r 2 - 16 



Figure 5.109 The capsule of 
Example 3. 




Figure 5.110 The shadow of 
the capsule in Figure 5.109 
in the z = 0 plane. 



In addition, the temperature function may be described in cylindrical coordi- 
nates as 



T(x,y,z) = z(x 2 + y 2 ) = zr 2 . 



Hence, we may calculate 



[^]avg — 



/o* io /2 ii-16 zrl -rdzdr dO 



ft ft 2 fl/-ib rdzdr de 



For the denominator integral, 

-lit r 5/2 r 9-r 2 



/ / / rdzdr d6 = / r ((9 - r 2 ) - (3r 2 - 16)) dr dd 

Jo J(i Jlr 2 -16 Jo Jo 

= / / (25r - 4r 3 ) dr de 
Jo Jo 

-f(i 

2n /625 625 



r 2 -r 4 



5/2 



II 



-f 

Jo 



625 625tt 

, de = 2jz = . 

8 16 7 16 8 



5.6 | Applications of Integration 



This result agrees with the volume calculation in Example 6 of §5.4, as it should. 
For the numerator integral, we compute 



p2n p5/2 p9-r 2 
JO Jo Jir 2 -U 



-2ji r 5/2 / .2 



zr dzdrd.9 



z=9-r 2 



dr dO 



z =3r 2 -16 



Thus, 



Jo Jo 

= / / ^{(9-r 2 ) 2 -(3r 2 -\6f)drde 
Jo Jo z 

f 2n [5/2 r 3 

= / / — (-8r 4 + 78r 2 - 175) dr dO 
Jo Jo 2 

J f2x p5/2 

= o/ / (-^ 7 + 78r 5 - I75r 3 )dr d9 
2 Jo Jo 

= 2/ 



_ r 8 + 13r 6 _ 121 r 4 

4 



2?r 15625 15625 

d6 = TV. 

256 256 



[^]avg — 



■156257T/256 25 



625^/8 



32 



Center of Mass: The Discrete Case 

Consider a uniform seesaw with two masses mj and m 2 placed on either end. If 
we introduce a coordinate system so that the fulcrum of the seesaw is placed at 
the origin, then the situation looks something like that shown in Figure 5.111. 
Note that x% < 0 < jci. The seesaw balances if 

m\X\ + mjX2 = 0. 

In this case, the center of mass (or "balance point") of the system is at the origin. 

But now suppose m\X\ + m 2 X2 ^ 0. Then where is the balance point? Let us 
denote the coordinate of the balance point by x. Before we find it, we'll introduce 
a little terminology. The product m, Xj (in this case, for i = 1, 2) of mass and 
position is called the moment of the ith body with respect to the origin of the 
coordinate system. The sum m 1X1 + 1112x2 is called the total moment with respect 
to the origin. To find the center of mass, we use the following physical principle, 
which tells us that a system of several point masses is physically equivalent (in 
terms of moments) to a system with a single point mass. 



Guiding physical principle. The center of mass is the point such that, if all 
the mass of the system were concentrated there, the total moment of the new 
system would be the same as that for the original system. 



Putting this principle into practice in our situation, we see that total mass M 
of our system is mi + m,2. If x is the center of mass, then the guiding principle 
tells that 



Mx = m\X\ + M2X2- 



Chapter 5 | Multiple Integration 



m 1 m 2 m 3 



X-i 



0 



Figure 5.112 A system of « 
masses distributed on a line. 



m l<t 



(*i>Vi) 



(*3>y 3 ) 



Figure 5.113 A system of « 
masses in R 2 . 



That is, the total moment of the new (concentrated) system is the same as the total 
moment of the original system. Hence, 

m\X\ + 1H2X2 



X = 



If we have a system of n masses distributed along a (coordinatized) line, then 
the same reasoning may be applied. (See Figure 5.1 12.) We have 



total moment ni\X\ + 1712X2 + • • • + m n x n Yl"=i m i x i 



total mass 



mi + m 2 H h m„ 



(2) 



Now we move to two and three dimensions. Suppose, first, that we have n 
particles (or bodies) arranged in the plane as in Figure 5.1 13. Then there are two 
moments to consider: 

n 

Total moment with respect to the y-axis = m,*, , 



1=1 



and 



Total moment with respect to the x-axis = m, -y,- . 



1=1 

(Admittedly, this terminology may seem confusing at first. The idea is that the 
moment measures how the system balances with respect to the coordinate axes. 
It is the x-coordinate — not the y-coordinate — that measures position relative to 
the y-axis. Similarly, the y-coordinate measures position relative to the x-axis.) 
The guiding principle tells us that the center of mass is the point (x , y) such that, 
if all the mass of the system were concentrated there, then the new system would 
have the same total moments as the original system. That is, if M = ^ m ; , then 



Mx = rriiXi 



(i.e., the moment with respect to the y-axis of the new system equals the moment 
with respect to the y-axis of the original system) and 



My = y^m,y,-. 
Thus, we have shown the following: 



Discrete center of mass in R 2 . Given a system of n point masses m\, 
m 2 , . . . , m„ at positions 

(xi,yi), (x 2 , y 2 ) (x„,y n ) inR 2 , 

the coordinates (x , y) of the center of mass are 



5.6 | Applications of Integration 379 

For particles arranged in three dimensions, little more is needed than adding 
an additional coordinate. (See Figure 5.1 14.) 



Discrete center of mass in R . Given a system of n point masses m\, 
ni2, ■ . ■ , m n at positions 

(x u yi,zi), (x 2 , y 2 , Zi), ■ ■ - , (x„,y„,z„) inR 3 , 

the coordinates (x, y, z) of the center of mass are given by 

*= , y= ^' — , and z= ^' ' ■ (4) 



The numerators of the fractions in (4) are the moments with respect to the 
coordinate planes. Thus, for example, the sum Yl"=i m i x > i s me tota l moment 
with respect to the yz-plane. 

By definition, moments of physical systems are additive. That is, the total 
moment of a system is the sum of the moments of its constituent pieces. However, 
it is by no means the case that a coordinate of the center of mass of a system is 
the sum of the coordinates of the centers of mass of its pieces. This additivity 
property makes the study of moments important in its own right. 

Center of Mass: The Continuous Case 

Now, we turn our attention to physical systems where mass is distributed in 
a continuous fashion throughout the system rather than at only finitely many 
isolated points. 

To begin with the one-dimensional case, suppose we have a straight wire 
placed on a coordinate axis between points x = a and x = b as shown in 
Figure 5.115. Moreover, suppose that the mass of this wire is distributed according 
to some continuous density function <5 (x ) . We seek the coordinate x that represents 
the center of mass, or "balance point," of the wire. 



x n =b 

Figure 5.1 1 5 A "coordinatized" wire. 
The mass of the segment between jc;_i and 
Xi is approximately 8(x*)Axj. 

Imagine breaking the wire into n small pieces. Since the density is continuous, 
it will be nearly constant on each small piece. Thus, for i = 1 , . . . , n, the mass m, 
of each piece is approximately 8(x*)Ax, ■ , where A the length of 

each segment of wire, and x* is any number in the subinterval [jc,_i , #,•]. Hence, 
the total mass is 

n n 

M = J2' n i ~ ^2s(x*)Axi, 

i=l i=l 

and the total moment with respect to the origin is approximately 

xf 8(x*)Ax,. 

approx. approx. 
position mass 




m i 



- y 



X 

Figure 5.1 1 4 A discrete system 
of masses in R 3 . 



J- — I 1 1 — H- 

d — Xq X\ X-j ■ ■ ■ X; _ i X; 



380 Chapter 5 | Multiple Integration 



Of course, these results can be used to provide an approximation of the coordi- 
nate x of the center of mass. For an exact result, however, we let all the pieces 
of wire become "infmitesimally small"; that is, we take limits of the foregoing 
approximating sums as all the Ax, 's tend to zero. Such limits give us integrals, 
and we may reasonably define our terms as follows: 



Continuous center of mass in R. For a wire located along the x-axis 
between x = a and x = b with continuous density per unit length <5(x): 



f 



Total mass = / 8(x)dx 



Total moment = / xS(x)dx. (5) 

J a 

total moment f xS(x)dx 

Center of mass x = 



total mass f* 8(x)dx 



Compare the formulas in (3) with those in (5). Instead of a sum of masses 
and a sum of products of mass and position, we have an integral of "infinitesimal 
mass" (the <5(x) dx term) and an integral of infinitesimal mass times position. 



EXAMPLE 4 Suppose that a wire is located between x = — 1 and x = 1 along 
a coordinate line and has density S(x) = x 2 + 1. Using the formulas in (5), we 
compute that the center of mass has coordinate 




Figure 5.116 A lamina 
depicted as a region D in 
the xy-plane with density 
function S. 



f\x{x 2 +\)dx _ {jx^+jx^ti 
f\(x 2 +l)dx (ix 3 +x)|lj 



4 = 0. 

3 



This makes sense, since this wire has a symmetric density pattern with respect to 
the origin (i.e., <5(x) = S(—x)). ♦ 

The analogous situation in two dimensions is that of a lamina or flat plate of 
finite extent and continuously varying density S(x, y). (See Figure 5.116.) Using 
reasoning similar to that used to obtain the formulas in (5), we make the following 
definition for the coordinates (x, y) of the center of mass of the lamina: 



Continuous center of mass in R 2 . For a lamina represented by the region 
D in the xy-plane with continuous density per unit area 8(x, y): 



x = 



total moment with respect to y-axis ff D x S(x, y)dA 

' Jf D S(x,y)dA ; 

ff D yS(x,y)dA 



total mass 

total moment with respect to x-axis 
total mass 



(6) 



ff D S(x,y)dA ■ 



5.6 | Applications of Integration 



Roughly, the term S(x, y)dA represents the mass of an "infinitesimal two-dimen- 
sional" piece of the lamina and the various double integrals the limiting sums of 
such masses or their corresponding moments. 

EXAMPLE 5 We wish to find the center of mass of a lamina represented by the 
region D in R 2 whose boundary consists of portions of the parabola y = x 2 and 
the line y = 9 and whose density varies as S(x, y) = x 2 + y. (See Figure 5.117.) 

First, note that this lamina is symmetric with respect to the y-axis and that, in 
addition, the density function has a similar symmetry because 8 (x, y) = S(—x, y). 
We may conclude from these two observations that the center of mass must occur 
along the y-axis (i.e., that x. = 0). Using the formulas in (6) and noting that the 
lamina is represented by an elementary region of type 1 , 



y = 



Jf D yS(x,y)dA _ j% f^yjx 2 + y)dy dx 
ff D S(x,y)dA : : /\ • y)dydx 



For the denominator integral, we compute 



x + y) dy dx = 




dx 






For the numerator, 




9 



y(x 2 + y)dy dx = 




dx 




x 6 x 6 \l 

1 dx 

2 3 )\ 



3 



11664 



7 



Hence, 



11664/7 _ 45 
1296/5 ~ T 



6.43. 



This answer is quite plausible, since the density of the lamina increases with y, 
and so we should expect the center of mass to be closer to y = 9 than to y = 0. 

♦ 



382 Chapter 5 | Multiple Integration 



We may modify the two-dimensional formulas to produce three-dimensional 
ones. 



Continuous center of mass in R 3 . Given a solid W whose density per 
unit volume varies continuously as 8(x, y, z), we compute the coordinates 
(x, y, z) of the center of mass of W using the following quotients of triple 
integrals: 

total moment with respect to yz-plane fff w x8(x,y,z)dV 
total mass 



fff w S(x,y,z)dV ' 



y 



total moment with respect to xz-plane fff w y K x i y, z)dV 



total mass fff w 8(x, y, z)dV 

total moment with respect to x y-plane fff w z S(x, y,z)dV 
total mass fff w S(x,y,z)dV 



-; (7) 



(0,0,3) 



Plane 

x + y + z =3 




(3,0, 0) 



Figure 5.118 

Example 6. 



The tetrahedron of 



In (7) we may think of the term S(x, y, z)dV as representing the mass of an 
"infinitesimal three-dimensional" piece of W. Then the triple integrals are the 
limiting sums of masses or moments of such pieces. 

EXAMPLE 6 Consider the solid tetrahedron W with vertices at (0, 0, 0), 
(3, 0, 0), (0, 3, 0), and (0, 0, 3). Suppose the mass density at the point (x, y, z) 
inside the tetrahedron is 8(x, y,z) = x + y + z+ 1. We calculate the resulting 
center of mass. (See Figure 5.118.) 

First, note that the position of the tetrahedron in space and the density function 
are both such that the roles of x, y, and z may be interchanged freely. Hence, the 
coordinates (x, y, z) of the center of mass must satisfy x — y — z. Therefore, we 
may reduce the number of calculations required. 

The tetrahedron is a type 4 elementary region in space. Thus, we may calculate 
the total mass M of W, using the following iterated integral: 



l-x r 3-x-y 



M 



if I 

Jo Jo Jo 
fo fo \ 
Jo Jo 

= f m- 

Jo 



(x + y + z + l)dzdy dx 

Z=3—x—y 



(x + y + \)z + 



1 T 2 

2 X 



y -xy 



dy dx 



2 y 2 ) dy dx 



x 2 ) y -\{\+x)y 2 -\y% 



1-2 

2 



=3-.i 



dx 



The total moment with respect to the xy-plane is given by 

O r 3-x i-3-x-y 



ITI 

Jo Jo Jo 



z(x + y + z + 1) dz dy dx 



5.6 | Applications of Integration 



•3 ,3-, / 2 z 3 

(x + y+l)- + j 



If 

Jo Jo 

/T"< 

Jo Jo 



=3-x-y 



dy dx 



2 2 X + + 



+ xy+ \x 2 y + \y 2 + \xy 2 + iy 3 ) dy dx 



Jo 



117 27, 



15 „2 _ 1 v 3 
4 A 6 A 



l x 4 )dx= 459 



Hence, 



x = y = z = 



459 
40 



117 



51 

65 



24 



0.7846. 



40 ■ 



If an object is uniform, in the sense that it has constant density, then one uses 
the term centroid to refer to the center of mass of that object. Suppose the object 
is a solid region W in R 3 . Then, if the density <5 is a constant k, the equations for 
the coordinates (x,y,z) may be deduced from those in (7). For the x-coordinate, 
we have 



x = 



fff w xS(x,y,z)dV = fff w kxdV 
fff w 8(x,y,z)dV fff w kdV 

fff w xdV _ 1 
fff w dV volume of W 



1 1 Iw 



xdV. 



Similarly, 
y = 



1 



volume of W 



f f fy 



ydV and z= 

volume of W 



ffl 



zdV. 



In particular, the constant density S plays no role in the calculation of the 
centroid, only the geometry of W. (Note: Completely analogous statements can 
be made in the case of centroids of laminas in R 2 .) 



Figure 5.1 19 The cone of 
Example 7. 



EXAMPLE 7 We compute the centroid of a cone of radius a and height h . (See 
Figure 5.119.) 

By symmetry, x = y = 0. Moreover, we know that the volume of the cone is 
(ir/3)a 2 h. Thus, the z-coordinate of the centroid is 



= —111 

na 2 h J J J w 



zdV. 



This triple integral is most readily evaluated by using cylindrical coordinates. 
(See Example 15 of §5.5.) The lateral surface of the cone is given by z = \r, so 
we calculate 



^ / , 2jr pa 

= — 7T- I I I zrdzdrd0 = 
na z h J 0 Jo J'i r 



Tca 2 h 



jta 2 h 2 \ 3 

I = -h 

4/4 



after a straightforward evaluation. Hence, the centroid of the cone is located at 
(0,0, \h). ♦ 



384 Chapter 5 | Multiple Integration 



Moments of Inertia 

Let W be a rigid solid body in space. As we have seen, the moment integral 
with respect to the xy-plane is M xy = fff w z 8(x, y, z)dV — that is, the inte- 
gral of the product of the position relative to a reference plane (in this case the 
xy-plane) and the density of the solid. This integral can be considered to mea- 
sure the ease with which W can be displaced perpendicularly from the reference 
plane. 

Now, consider spinning W about a fixed axis (which may or may not pass 
through W). The moment of inertia / (or second moment — the moment integral 
mentioned in the preceding paragraph is sometimes called the first moment) is 
a measure of the ease with which W can be made to spin about the given axis. 
Specifically, / is the integral of the product of the density at a point in W and the 
square of the distance from that point to a fixed axis; that is, 



1 1 Iw 



d l 8{x,y,z)dV, (8) 

w 

where d is the distance from (x, y, z) 6 W to the specified axis. 
When the axes of rotation are the coordinate axes in R 3 , we have 



/ v = moment of inertia about the x-axis = / / / (y + z )8(x, y, z) dV; 

'w 



ffl 

I y = moment of inertia about the y-axis = / / / (x 2 + z 2 ) S(x, y, z) d V; 

J J Jw 

I- = moment of inertia about the z-axis = / / / (x 2 + y 2 )S(x, y, z)dV. 

J J Jw 




Figure 5.120 The box of 

Example 8. 



EXAMPLE 8 Let W be a solid box of uniform density S and dimensions a, b, 
and c. If W is situated symmetrically with respect to the coordinate axes as shown 
in Figure 5.120, we compute the moments of inertia with respect to these axes. 
Note, first, that W may be described as 



W = \(x,y,z) 



a a b be c 

- < x < -, — < y < — < z < - 
2 ~ ~ 2 2~ y ~ 2 2 ~ "2 



Hence, the moment of inertia about the x-axis is 

/c/2 rb/2 pa /2 pc/2 pb/2 

/ / (v 2 + z 2 )Sdxdydz= / / (y 2 + z 2 )Sadydz 
-c/2 J-b/2 J -a/2 J -c/2 J-b/2 

f c/2 / 3 \ y= b l 2 f c /2 / b 3 \ 

= Sa LS^ +zy ),---J z=Sa LM +bz ) 



+ bz z dz 



,'b^c bc^ 

Sal 1 

12 12 



Sabc 



12 



'-(b 2 + c 2 ). 



5.6 | Applications of Integration 



By permuting the roles of x, y, and z (and the corresponding constants a, b, and 
c), we see that 

Sabc 0 i Sabc , , 

l y = —{a 2 + c 2 ) and L = —{a 2 + b 2 ). 

Therefore, if a > b > c (as in Figure 5.120), it follows that I x < I y < /,. This 
result may be confirmed by the observation that rotations about the axis parallel 
to the longest side of the box are easiest to effect in that the same torque applied 
about each axis will cause the most rapid rotation to occur about the axis through 
the longest dimension. A related fact is regularly exploited by figure skaters who 
pull their arms in close to their bodies, thereby reducing their moments of inertia 
and speeding up their spins. ♦ 

EXAMPLE 9 Let W be the solid bounded by the cone z = 2^/x 2 + y 2 and the 
plane z = 4 shown in Figure 5.121. Assume that the density of material inside W 
varies as 8(x, y, z) = 5 — z. Let us calculate the moment of inertia I z about the 
z-axis. 

Given the geometry of the situation, it is easiest to work in cylindrical coor- 
dinates, in which case the cone is given by the cylindrical equation z = 2r. Thus, 
we have 

L = f [ [ (x 2 + y 2 )S(x,y,z)dV = f f f r 2 (5 - z)rdzdrd6 

J J Jw Jo Jo Jlr 

= / r3 ( 5z ~ \ z2 )\ Z ~2r drM = / / ( 12r3 " 10r 4 + 2r 5 ) drde 
Jo Jo Jo Jo 

f 2n , i s 1 /r\ i r=2 f 2lT 16 32;r 

Jo Jo 



Recall that the center of mass of a solid object of total mass M is the point 
such that if all the mass M were concentrated there, the (first) moment would 
remain the same. An analogous idea may be defined in the context of moments of 
inertia. The radius of gyration of a solid with respect to an axis is the distance 
r from that axis that we should locate a point of mass M so that it has the same 
moment of inertia / (with respect to the axis) as the original solid does. More 
concisely, the radius of gyration r is defined by the equation 




EXAMPLE 10 We determine the radius of gyration with respect to the z-axis 
of the cone described in Example 9. Hence, we compute 



386 Chapter 5 | Multiple Integration 



From Example 9, L = 32n/3. We determine the total mass M of the cone as 
follows: 

M= / / (5 - z)rdzdrd6 = / (12 - lOr + 2r 2 )r dr d9 

J0 J0 Jlr Jo Jo 



=r((* i -^+Mi^*=r^=f- 

Thus, 

' Y 32^/3 



5.6 Exercises 



1 . The local grocery store receives a shipment of 75 cases 
of cat food every month. The inventory of cat food (i.e., 
the number of cases of cat food on hand as a function 
of days) is given by I(x) = 75 cos(jrx/15) + 80. 

(a) What is the average daily inventory over a month? 

(b) If the cost of storing a case is 2 cents per day, 
determine the average daily holding cost over the 
month. 

2. Find the average value of f(x, y) = sin 2 x cos 2 y over 
R = [0,2jt] x [0,4tt]. 

3. Find the average value of f(x, y) = e 2x+y over the tri- 
angular region whose vertices are (0, 0), (1, 0), and 
(0, 1). 

4. Find the average value of g(x, y, z) = e z over the unit 
ball given by 

B = {(x,y,z) | * 2 + y 2 + z 2 < 1}. 

5. Suppose that the temperature at a point in the cube 

W = [-1, 1] x [-1, 1] x [-1, 1] 

varies in proportion to the square of the point's distance 
from the origin. 

(a) What is the average temperature of the cube? 

(b) Describe the set of points in the cube where the 
temperature is equal to the average temperature. 

6. Let D be the region between the square with ver- 
tices (1, 1), (-1, 1), (-1, -1), (1, -1) and the unit 
disk centered at the origin. Find the average value of 
f(x, y) = x 2 + y 2 on D. 

7. Let W be the region in R 3 between the cube with 
vertices (1, 1, 1), (-1, 1, 1), (-1, -1, 1), (1, -1, 1), 
(1,1,-1), (-1,1,-1), (-1,-1,-1), (1,-1,-1) 
and the unit ball centered at the origin. Find the av- 
erage value of f(x, y, z) = x 2 + y 2 + z 2 on W. 



8. Suppose that you commute every day to work by sub- 
way. You walk to the same subway station, which is 
served by two subway lines, both stopping near where 
you work. During rush hour, each subway line sends 
trains to arrive at the stop every 6 minutes, but the dis- 
patchers begin the schedules at random times. What is 
the average time you expect to wait for a subway train? 
(Hint: Model the waiting time for the two subway lines 
by using a point (x, y) in the square [0, 6] x [0, 6].) 

9. Repeat Exercise 8 in the case that the subway stop is 
serviced by three subway lines (each with trains arriv- 
ing every 6 minutes), rather than two. 

10. Find the center of mass of the region bounded by the 
parabola y = 8 — 2x 2 and the *-axis 

(a) if the density S is constant; 

(b) if the density S = 3y. 

11. Find the centroid of a semicircular plate. (Hint: Judi- 
cious use of a suitable coordinate system might help.) 

1 2. Find the center of mass of a plate that is shaped like the 
region between y = x 2 and y = 2x, where the density 
varies as 1 + x + y. 

13. Find the center of mass of a lamina shaped like the 
region 

{(*, y) | 0 < y < *Jx, 0 < x < 9}, 
where the density varies as xy. 

1 4. Find the centroid of the region bounded by the cardioid 
given in polar coordinates by the equation r = 
1 — sin 9. (Hint: Think carefully.) 

15. Find the centroid of the lamina described in polar 
coordinates as 

{(r,9) | 0 < r < 4cos0, 0 < 6 < tt/3}. 



5.6 I Exercises 387 



1 6. Find the center of mass of the lamina described in polar 
coordinates as 

l(r,6) | 0 < r < 3, 0 <6 < tt/4}, 

where the density of the lamina varies as S(r, 0) = 
4-r. 

17. Find the center of mass of the region inside the car- 
dioid given in polar coordinates as r = 1 + cos 8, and 
whose density varies as S(r, 6) = r. 

1 8. Find the centroid of the tetrahedron whose vertices are 
at (0, 0, 0), (1, 0, 0), (0, 2, 0), and (0, 0, 3). 

19. A solid is bounded below by z = 3y 2 , above by the 
plane 2 = 3, and on the ends by the planes x = — 1 
and x = 2. 

(a) Find the centroid of this solid. 

(b) Now assume that the density of the solid is given 
by S = z + x 2 . Find the center of mass of the 
solid. 

20. Determine the centroid of the region bounded above 
by the sphere x 2 + y 2 + z 2 = 1 8 and below by the 
paraboloid 3z = x 2 + y 2 . 

21 . Find the centroid of the solid, capsule-shaped region 
bounded by the paraboloids z = 3x 2 + 3y 2 — 16 and 

z = 9-x 2 -y 2 . 

22. Find the centroid of the "ice cream cone" shown in 
Figure 5.122. 



z 

Sphere: x 2 + y 2 + z 2 = 25 




x 



Figure 5.122 The ice cream cone solid 
of Exercise 22. 

23. Find the centroid of the solid shaped as one-eighth of 
a solid ball of radius a. (Hint: Model the solid as the 
first octant portion of a ball of radius a with center at 
the origin. ) 

24. Find the center of mass of a solid cylindrical peg of 
radius a and height h whose mass density at a point 
in the peg varies as the square of the distance of that 
point from the top of the cylinder. 



25. (a) Find the moment of inertia about the coordinate 

axes of a solid, homogeneous tetrahedron whose 
vertices are located at (0,0,0), (1, 0, 0), (0, 1, 0), 
and(0, 0, 1). 

(b) What are the radii of gyration about the coordinate 

axes? 

26. Consider the solid cube W = [0, 2] x [0, 2] x [0, 2]. 
Find the moments of inertia and the radii of gyration 
about the coordinate axes if the density of the cube is 
S(x, y, z) = x + y + z + 1. 

27. A solid is bounded by the paraboloid z = x 2 + y 2 and 
the plane z = 9. Find the moment of inertia and radius 
of gyration about the z-axis if 

(a) the density is S(x, y, z) = 2z; 

(b) the density is S(x, y, z) = *Jx 2 + y 2 . 

28. Find the moment of inertia and radius of gyration about 
the z-axis of a solid ball of radius a, centered at the ori- 
gin, if 

(a) the density S is constant; 

(b) S(x,y,z) = x 2 + y 2 + z 2 ; 

(c) S(x,y,z) = x 2 + y 2 . 

We can find the moment of inertia of a lamina in the plane 
with density S(x, y)by considering the lamina to be a flat plate 
sitting in the xy -plane in R 3 . Then, for example, the distance 
of a point (x, y) in the lamina to the x-axis is given by \y\, the 
distance to the y-axis is given by \x\, and the distance to the 

z-axis (or the origin) is given by *J x 2 + y 2 . (See Figure 5.123.) 
Using these ideas, find the specified moments of inertia of the 
laminas given in Exercises 29—31. 



z 




29. The moment of inertia I x about the x-axis of the 
lamina that has the shape bounded by the graph of 
y = x 2 +2 and the line y = 3, and whose density 
varies as S(x, y) = x 2 + 1. 

30. The moment of inertia /, about the z-axis of the lam- 
ina shaped as the rectangle [0, 2] x [0, 1], and whose 
density varies as S(x, y) = 1 + y. 



388 Chapter 5 | Multiple Integration 



31 . The moment of inertia about the line y = 3 of the lam- 
ina shaped as the disk 

{(x,y)\x 2 + y 2 <4}, 

and whose density varies as S(x, y) = x 2 . 

The gravitational field between a mass M concentrated at 
the point (x, y, z) and a mass m concentrated at the point 
(xo, Jo, Zo) is 

p _ GMm[(x - x 0 )i + (y - ypjj + (z - z 0 )k] 
[(x - x 0 ) 2 + (y - y 0 ) 2 + (z - z a ) 2 fl 2 
The gravitational potential V ofF is 

_ GMm 

y/(x -x 0 ) 2 + (y- y 0 ) 2 + (z - Z 0 ) 2 ' 

(We have seen in §3.3 that F = -VV.J Now suppose that, in- 
stead of a point mass M, we have a solid region W of density 
S(x, y, z) and total mass M. Then the gravitational potential of 
W acting on the point mass m may be found by looking at "in- 
finitesimal "point masses dm = S(x, y,z)dV and adding (via 
integration) their individual potentials. That is, the potential 
ofW is 

V(x 0 , y 0 , z 0 ) 

err GmS(x, y , z )dv 

J J Jw J{x - x 0 ) 2 + (y - y 0 ) 2 + (z - z 0 ) 2 ' 

In Exercises 32-34, Let W be the region between two con- 
centric spheres of radii a < b, centered at the origin. (See 
Figure 5.124.) Assume that W has total mass M and constant 
density 8. The object of the following exercises is to compute 
the gravitational potential V(xo, yo, Zo) of W on a mass m 
concentrated at (xo, yo, Zo)- Note that, by the spherical sym- 
metry, there is no loss of generality in taking (xq, yo, Zo) equal 



to (0, 0, r). So, in particular, r is the distance from the point 
mass m to the center ofW. 




Figure 5.124 The spherical 
shell of Exercises 32-34. 



32. Show that if r > b, then V(0, 0, r) = -GMm/r. This 
is exactly the same gravitational potential as if all the 
mass M of W were concentrated at the origin. This is a 
key result of Newtonian mechanics. (Hint: Use spher- 
ical coordinates and integrate with respect to <p before 
integrating with respect to p.) 

33. Show that if r < a, then there is no gravitational force. 
(Hint: Show that V(0, 0, r) is actually independent of r. 
Then relate the gravitational potential to gravitational 
force. As in Exercise 32, use spherical coordinates and 
integrate with respect to (p before integrating with re- 
spect to p.) 

34. (a) Find V(0, 0, r) if a < r < b. 

(b) Relate your answer in part (a) to the results of Ex- 
ercises 32 and 33. 



5.7 Numerical Approximations of Multiple 
Integrals (optional) 

Suppose that / is a continuous function defined on some bounded region D in the 
plane. If D is a rectangle or an elementary region, then Theorem 2. 1 0 enables us to 
calculate ff D f dA as an iterated integral. However, the "partial antiderivatives" 
in the iterated integral may turn out to be impossible to determine in practice. 
Nonetheless, an approximate value of ff D f dA may be entirely acceptable for 
some purposes. In this section, we discuss how we can adapt numerical meth- 
ods for approximating definite integrals of functions of a single variable to give 
techniques for approximating double integrals. 

Numerical Methods for Definite Integrals 

Let us review two familiar techniques for approximating the value of f a f(x) dx. 
The first of these is the trapezoidal rule. We begin by partitioning the interval 
[a, b] into n equal subintervals. Thus, we set 

b — a 

Ax = , a = xo < x\ < . . . < x n = b, where Xj = a + i Ax. 

n 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



The trapezoidal rule approximation T„ is 

T n = ^ U(a) + 2 fix,) +■■■+ 2/(x»_i) + f(b)] 

(1) 

It is obtained by approximating the function / with n linear functions that pass 
through the pairs of points (x,_i , /(x,_i)) and (x,- , /(*,•)) for r = 1, . . . , n. Then 
the (net) area under the curve y = f(x) between x = a andx = b is approximated 
by the sum of the (net) areas under the graphs of the linear functions. When / 
is nonnegative, the approximating areas are those of trapezoids — thus, the name 
for the method. (See Figure 5.125.) 

The key theoretical result concerning the trapezoidal rule is given by the 
following. 



Ax 
~2~ 



n-l 



/(a) + 2 £/(*,) + /(&) 



THEOREM 7.1 Given a function / that is integrable on [a, b], we have 



/' 

J a 



f{x)dx = T n + E„, 



where 



Ax 
~1~ 



n-l 



f(a) + 2j2f(xi) + f(t>) 



and E„ denotes the error involved in using T n to approximate the value of the 
definite integral. In addition, if / is of class C 2 on [a, b], then there is some 
number f in (a , b) such that 



F — 



(Ax) 2 /"(0 = 



12 



12n 2 



In particular, Theorem 7.1 shows that if /" is bounded that is, if \f"(x)\ < M 
for all x in [a, b], then \E n \ < (b — a) 3 M/(l2n 2 ). This inequality is very useful 
in estimating the accuracy of the approximation. 

EXAMPLE 1 We approximate f 0 sin(x 2 ) dx using the trapezoidal rule with 
n = 4. Thus, we have 

1 - 0 

Ax = = 0.25, 

4 

so that, using (1), we have 



I 



1 0 25 

sin(x 2 )Jx R« T 4 = — [sin0 + 2sin(0.25 2 ) + 2sin(0.5 2 ) + 2sin(0.75 2 ) 



+ sinl] 
0.25 



2 
0.25 



[0 + 0.12492 + 0.49481 + 1.06661 + 0.84147] 
[2.52780] = 0.31598. 



Chapter 5 | Multiple Integration 



Note that the second derivative of sin(x 2 ) is 2 cos (x 2 ) — Ax 2 sin(x 2 ) so that, for 
0 < x < 1, 

|2cos(x 2 )-4x 2 sin(x 2 )| < 2| cos(x 2 )| + 4x 2 | sin(x 2 )| < 2 + 4 = 6. 
Hence, using Theorem 7.1, 

(1 - 0) 3 6 1 

|£ 4 | < - V- = — = 0.03125. 

1 1 " 12 -4 2 32 

Thus, the true value ofthe integral must be between 0.3 1598 - 0.03125 = 0.2847 
(rounding to four decimal places) and 0.31598 + 0.03125 = 0.3473. Of course, 
a more accurate approximation may be obtained by taking a finer partition (i.e., 
a larger value for n). ♦ 

Another numerical technique for approximating the value of f(x)dx with 
which you may be familiar is Simpson's rule. As with the trapezoidal rule, we 
partition the interval [a, b] into equal subintervals, only now we require that the 
number of subintervals be even, which we will write as 2n. Thus, we take 

b — a 

Ax = , a = xo < X\ < . . . < X2 n = b, where x,- = a + / Ax. 

2n 

Then the Simpson's rule approximation Sin is 

Ax 

Sm = — If (a) + 4/(xi) + 2/(x 2 ) + 4/(x 3 ) + • ■ ■ + 2/(x 2 „_ 2 ) 
+ 4/(x 2n _0 + /(&)] 

n — 1 n 

f(a) + 2 £ /(x 2( ) + 4 £ /(x 2l _.) + f(b) 



Ax 



(2) 



EXAMPLE 2 We approximate the value of f 0 sin(x 2 ) dx of Example 1 using 
Simpson's rule with n = 2 (i.e., four subintervals). As before, we have 

1 - 0 

Ax = = 0.25, 

4 

so (2) gives 

■! o 25 

sin (x 2 ) dx R» 5 4 = — [sin 0 + 4 sin (0.25 2 ) + 2 sin (0.5 2 ) + 4 sin (0.75 2 ) 



/ 

Jo 



+ sin 1] 
0.25 



3 

0.25 



[0 + 0.24984 + 0.49481 + 2.13321 + 0.84147] 
[3.71933] = 0.30994. 



3 

Note that this value is in the range predicted by the trapezoidal rule. In fact, it is 
a more accurate approximation to the value of the definite integral. ♦ 

Simpson's rule is obtained by approximating the function / with n quadratic 
functions that pass through triples of points (x 2 ,_ 2 , /(x 2( _ 2 )), (x 2 ;_i, /fe-i)), 
(x 2i , f(xii)) for i = 1 , . . . , n. As with the trapezoidal rule, we have the following 
summary result. 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



THEOREM 7.2 Given a function / that is integrable on [a, b], we have 

b 

f(x)dx = S 2n + E 2n , 
where 

n— 1 n 

f(a) + 2 £ /(xa) + 4 J] /(x 2i _0 + /(*) , 
i=i i=i 

and Ein denotes the error involved in using S^i to approximate the value of the 
definite integral. In addition, if / is of class C 4 on [a, b], then there is some 
number £ in (a , b) such that 

E, - - b ~ a (Ax) 4 f^>(f) - - (fc - a) f (4) m 
2 " " 180 { ' 1 (;) ~ 2880/2 4 1 {;) - 



L 



Sin = 



Ax 



EXAMPLE 3 Consider (x 3 + 3x 2 ) dx. We compare the trapezoidal rule 
and Simpson's rule approximations with 4 subintervals. We thus have Ax = 
(3 - l)/4 = 0.5 and 

0.5 

T 4 = — [4 + 2(10.125 + 20 + 34.375) + 54] = 46.75; 
0.5 

S 4 = — [4 + 4(10.125) + 2(20) + 4(34.375) + 54] = 46. 

Note that 

(x 3 + 3x 2 ) dx = (I* 4 + x 3 ) \\ = (f + 27) - (i + 1) = 46, 



so Simpson's rule agrees with the exact answer. This should not be a surprise, 
since for f(x) = x 3 + 3x 2 , we have that f {4 \x) is identically zero, which means 
that the error term £4 for Simpson's rule must be zero. ♦ 

Approximating Double Integrals over Rectangles 



Now let / be a function of two variables that is continuous on the rectangle 
R = [a, b] x [c, d] in R 2 . We adapt the previous ideas to provide methods for 
approximating the value of the double integral Jj R fdA. 

Because we assume that / is continuous on R, Fubini's theorem applies to 

give 

/ / f(x,y)dA= f f f(x,y)dydx. (3) 

J J R J a Jc 

Next we partition the x-interval [a, b] into m equal subintervals. Thus, 
b — a 

Ax = , a = xo < x\ < . . . < x m = b, where x, ■ = a + / Ax. 

m 

Similarly, we partition the y-interval [c, d] into n equal subintervals; hence, 
d — c 

Ay = , c = yo < v\ < . . . < y„ = d, where y ; - = c + j Ay. 

n 

In the inner integral f d f(x, y) dy of the iterated integral in (3), the variable x is 
held constant. Therefore, we may approximate this integral using the trapezoidal 



Chapter 5 i Multiple Integration 



rule of Theorem 7.1. We obtain 



f{x,y)dy 



Ay 



n-l 



/(ac,c) + 2^)/(x,y;) + /(x,d) 

7 = 1 



Next, integrate each function of x appearing on the right side, so that 

-b pel 



J a J c 



Ay 
2 



y)dydx & 

r-b n — 1 pb r-b 

J f{x, c)dx + 2 I fi x ^yj)dx+ I fix,d)dx 

J a ,-_ i J a J a 



(4) 



Now use the trapezoidal rule again on each integral appearing on the right. This 
means that, for j = 0, . . . , n, 



f 

J (I 



fix, yj)dx 



Ax 



f(a t yj) + 2j2f(Xi,yj) + m yj) 



(5) 



Putting (4) and (5) together, we obtain 



p b pel 

/ / /(*, 

J a J c 



y) dy dx 



Ay Ax 



m— 1 



n-l 



Ax 



fia,c) + 2 Y J fi.x i ,c) + fib, c) 

m— 1 

f(a.y,) • 2j./|v,.v / ) • /(/>. V;) 



+ 



7=1 

Ax 



i=i 



fia,d) + 2Y J f(x i ,d) + fib,d) 



Therefore, the trapezoidal rule approximation T„ hn to ff R f dA is 



T 

1 m.n 



Ax Ay 



-l 



fia,c) + 2j2f(xi,c) + fQ>,c) 



( = 1 

n— 1 m — 1 



n— 1 n— 1 m— 1 ft— 1 

+ 2 J] /(a, y,) + 4££ /(x, , yj ) + 2 J] fib, yj) (6) 

7 = 1 7 = 1 «=1 7 = 1 



1-1 



+ fia,d) + 2Y J f{x i ,d) + fib,d) 



i=i 



The expression appearing in (6) is not too memorable as it stands. However, 
we may interpret it as follows: 



Ax Ay -r-^ ^-^ 

T m ,„ = —— 2^ yj)> 

j=0 1=0 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



where 



1 if (xi , yj) is one of the four vertices of R ; 

2 if (xi , yj) is a point on an edge of R, but not a vertex; 
4 if (x, , yj) is a point in the interior of R. 



EXAMPLE 4 We approximate the value of L j x {xy + 3x)dydx with 7^2. 
Thus, the ^-interval [3, 6] is partitioned using Ajc = (6 — 3)/3 = 1 and the y- 
interval [1, 2] is partitioned using Ay = (2 — l)/2 = 0.5. See Figure 5.126 for 
the rectangle [3,6] x [1,2] with partition points marked. 
Hence, we have 



(xy + 3x) dy dx « 
1(0.5) 



[12 + 24 + 30+ 15 



1 



+ 2(16 + 20 + 27 + 20 + 25 + 13.5) + 4(18 + 22.5)] 
-(486) = 60.75. 



y 

2 -- 
1.5 
1 + 



15 

13.5, 
12 



20 


25 


30 


18 


22.5 


27 


• 


• 




16 


20 


24 



H 1 1 h 



H h 



Figure 5.126 The rectangle 
[3, 6] x [1, 2] of Example 4 with 
partition points marked with the 
values of f(x, y) = xy + 3x. 



For comparison, we calculate the iterated integral directly: 

2 

dx 

y=l 



f 6 9 9 2 

= / -x dx = -x 
J 3 2 4 

Note that T3 2 gives the exact answer in this case. ♦ 

In the derivation above of the trapezoidal rule we assumed that the func- 
tion / was continuous. Nonetheless, we may use formula (6) to approximate 
ff R f dA whenever / is integrable on R. However, in order to make estimates 
of the accuracy of trapezoidal approximations, we must assume more about /, as 
the following result (whose proof involves use of the intermediate value theorem 
and the mean value theorem for integrals) indicates. 



if; 



(xy + 3x)dy dx 



= f( 



xy- 
2 



3xy 



243 



= 60.75. 



394 Chapter 5 | Multiple Integration 



THEOREM 7.3 (TRAPEZOIDAL RULE FOR DOUBLE INTEGRALS OVER RECTANGLES) 

Given a function / that is integrable on the rectangle R = [a , b] x [c, d], we have 



//. 



fix, y)dA = T m%n + E m>n , 

R 

where T„ u „ is given by (6) and E mj , denotes the error involved in using T„ un to 
approximate the value of the double integral. Moreover, if / is of class C 2 on R, 
then there exist points (fi , rj\) and (£2, Vi) in R such that 

E m , n = _ ( fc -°X^- c ) [(A*) 2 /^, , x ) + (Av) 2 / yy fe, ift)] • 



EXAMPLE 5 For /(*, y) = xy + 3x, we have that d 2 f/dx 2 and d 2 f/dy 2 are 
both identically zero. Hence, Theorem 7.3 implies that the trapezoidal rule ap- 
proximation is exact, as we already observed in Example 4. ♦ 

In a manner entirely analogous to our derivation of the trapezoidal rule, we 
may also produce a Simpson's rule approximation to the value of the double 
integral ff R f dA. As in the case of Simpson's rule for approximating single- 
variable definite integrals, we partition both the x- and y-intervals into even 
numbers of equal subintervals. Thus, we take 

b — a 

Ax = , a = xq < x\ < . . . < X2 m = b, where x,- = a + / Ax, 

2m 

and 

d — c 

Ay = , c = yo < y\ < . . . < ym = d, where y 7 - = c + j Ay. 

2n 

The resulting Simpson's rule approximation S2 m ,2n is given by the expression 



Ax Ay 



m— 1 m 

fia, c) + 2j2 f(*2i ,c) + 4J2 f(.X2i-u c) + fib, c) 

(=1 (=1 

— 1 n— 1 m— 1 n— 1 m 



+ 2 f(a, yij) f(*»> *y) + 8 E E 

;=1 7=1 i=l ;=1 t=l 

n-1 

+ 2J2fib,y 2 j) (7) 

7 = 1 

+ fia, yij-x) + /(**. yy-i) 

7=1 ;=1 i=l 

n m n 

+ 16 E E /c^-i. ^-i) + 4 E ^-0 

7=1 i=l 7=1 

m — 1 m 

+ /(fl, J) + 2 J] /(x 2 „ rf) + 4£ f(x2i-i,d) + fib, d) 
/=! i=i 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



Just as in the case of the trapezoidal rule for double integrals over rectangles, the 
intermediate value theorem and the mean value theorem for integrals provide the 
following result. 



THEOREM 7.4 (Simpson's rule for double integrals over rectangles) 
Let / be a function that is integrable on the rectangle R = [a, b] x [c, d]. Then 




f(x, y)dA = S2,„,2n + ^2m,2n> 



where S2m,2n is given by (7) and E 2m ,2„ denotes the error involved in using S% m ^ n 
to approximate the value of the double integral. Moreover, if / is of class C 4 on 
R, then there exist points (£i , r]\) and (£2, m) in R such that 

E 2m ,2n = ~ (b ~ a ^ K AX ) 4 + (Ay)Vvvvv(£2, m)] ■ 



EXAMPLE 6 We compare approximations to the value of / 0 °' 5 /„' e x+y dydx 
using both the trapezoidal rule and Simpson's rule with four subintervals in 
each of the x- and y-intervals. Thus, the x-interval [0, 0.5] is partitioned using 
Ax = (0.5 — 0)/4 = 0.125 and the y-interval [0, 1] is partitioned using Ay = 
(1 -0)/4 = 0.25. 

The trapezoidal rule gives 

T 4A = ( Q - 125 X°- 25 > yO+0 + e 0+0.5 + e l+0 + e 0.5+l + 2 ^0+0.25 + e 0+0.5 
_|_ ^0+0.75 _j_ e 0.125+0 _j_ ^0.25+0 _|_ ^0.375+0 _|_ ^0.125+1 _|_ ^0.25+1 



_|_ ^0.375+1 _|_ ^0.5+0.25 _j_ e 0.5+0.5 _j_ ^0.5+0.75^ 

0.125+0.25 | „0.125+0. 5 _|_ g 0. 125+0.75 _|_ g 0.25+0.25 
_|_ g 0.375+0.5 + ^0.375+0.75^ j 



+ 4(e" 

, ^0. 25+0. 75 , g 0.37 



1 

= (143.608854) = 1.121944. 

128 

Note that d 2 /dx 2 (e x+y ) = d 2 /dy 2 (e x+y ) = e x+y . Theorem 7.3 says that there 
exist points rji) and (£2, V2) in the rectangle [0, 0.5] x [0, 1] such that 

Eaa = _ (0-5-0)(l -0) ^ (0 _ 125) 2 efl+1J1 + (0 . 2 5)V 2 +" 2 ] . 

On the rectangle [0, 0.5] x [0, 1], the smallest possible value of e x+y is e 0+0 = 1 
and the largest possible value is e 0 5+l = e LS . Hence, we must have 

_(Ogl) j- (0 125) 2 e i. 5 + (0.25)2^-5] < e 4A < [ (0 . 125 ) 2 + (0.25) 2 ] 

or 

-0.0145888 < £44 < -0.003255. 



Chapter 5 | Multiple Integration 



|e 0+0.25 + ^0+0.75 + g 0.125+0 + ^0.125+1 + ^0.25+0.5 
g 0.375+0 . g 0.375+l . g 0.5+0.25 , ^0.5+0.75-, 



Hence, the true value of the double integral lies between 1.121944 — 0.0145888 = 
1.10735 and 1.121944 - 0.003255 = 1.11869. 
On the other hand, Simpson's rule gives 

5 4 ,4 = (0 - 125 ^ (0 - 25) + + e H0 + ^0.5+1 

+ 2 ( e °+ 0 - 5 + e °- 25+0 + e °- 25+1 + e °- 5+0 - 5 ) 
+ 4(e 

" 5 ) 

_|_ g ^0.125+0.5 _|_ ^0.25+0.25 _|_ g 0.25+0.75 _|_ g 0.375+0.5^ 

_|_ ! g ^0.125+0.25 _|_ ^0.125+0.75 _|_ ^0.375+0.25 _|_ ^0.375+0.75^ 

= —(321.036910) = 1.114711. 
288 

In this case, we note that d 4 /dx 4 (e x+y ) = d 4 /dy 4 (e x+y ) = e x+y , so, as before, 
the minimum and maximum values of these partial derivatives on [0, 0.5] x [0, 1] 
are, respectively, 1 and e . Therefore, Theorem 7.4 implies that 

[(°- 125 ) V - 5 + (0-25)V- 5 ] < £ 4 .4 < [(0.125) 4 + (0.25) 4 ] 

or 

-0.0000516688 < £ 4 ,4 < -0.0000115289. 

Hence, the true value of the double integral lies between 1.114711 — 
0.0000516688 = 1.11466 and 1.114711 - 0.0000115289 = 1.1147. 
For comparison, we may calculate the iterated integral exactly: 

/ / e x+y dydx=\ (e x+y )f dx = / (e x+l - e x ) dx 
Jo Jo Jo ' Jo 

= (e x+l - e x )\° Q 5 = e hS - e 0 - 5 - e + \ ^ 1.114686. 

Thus, we see that Simpson's rule gives a highly accurate approximation with a 
very coarse partition. ♦ 



Approximating Double Integrals over Elementary Regions 

We can modify the methods for approximating double integrals over rectangles 
to approximate double integrals over more general regions. Suppose first that D 
is an elementary region of type 1 ; that is, 

D = {{x, y) 6 R 2 I y(x) < y < S(x), a <x <b}. 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



Then, if / is continuous on D, Theorem 2.10 tells us that 

Iy(x) 



fdA= / f(x,y)dydx. 

J JD J a Jy(x) 



To approximate this iterated integral with a version of, say, the trapezoidal rule, 
we need to partition the region D in a reasonable way. We do so as follows. First, 
we partition the x -interval [a, b] in the usual way: 

b — a 

Ax = , a = xo < x\ < . . . < x m = b, where x l ■ = a + iAx. 

m 

Now, for a fixed x in [a, b], we partition the corresponding y-interval [y(x), S(x)] 
into n equal subintervals: 

. . , <5(x) - y(x) 
Ay(x) = , 



Y(x) = yo < y\ < ■ ■ ■ < y n = §(x), where yj(x) = y(x) + JAy(x). 

Note that now Ay and the partition numbers yo , . . . , y„ are all functions of x . (See 
Figure 5.127.) Then, by applying the trapezoidal rule first to the inner integral 
and then the outer one, we obtain 



pb pS(x) 
J a J y(x ) 



fix, y)dy dx 



L 



» Ay(x) 



n-l 



f(x, yix)) + 2^/(x, yjix)) + fix, Six)) 



dx 



Ax 
~1~ 



Ay id) ^— \ Ay(x, ) 
-——fia, y(a)) + 2 ^ — —fixi, y(x,)) 



Ay(fe) 



n-l 



——fia, yjia)) + 2 ^ — - — y 7 - (*,■)) 



+ -± 2 f(b,y j ib)) 



+ 



+ 



2 

Ay (a) 
2 

Ay(&) 



m— 1 



/(a, 5(a)) + 2 J] /(*,-, «(*,)) 



398 Chapter 5 | Multiple Integration 

The trapezoidal rule approximation is, thus, 



T = 



Ax Ay (a) 



4 



-i-i 



f(a, y(a)) + 2 J] f(a, yj(a)) + f(a, 8(a)) 



AxAy(xi) 



n-l 



2f( Xi , y(xt)) + f(*i , (8) 

7 = 1 



+ 2/(x i ,<5(x i )) 



+ 



AxAy(Z?) 



n-l 



y(6)) + 2 J] /(ft, yj (b)) + /(&, 5(6)) 



EXAMPLE 



4 

3.5 
3 
2.5 

2 



2 2.1 2.2 

Figure 5.1 28 The partitioned 
region D of Example 7. 



7 We approximate / 2 2 ' 2 J" 2 * (x 3 + y 2 ) 6?y dx by T 2i 4. We have 
2.2-2 

Ax = = 0.1, so xo = 2, xi = 2.1, X2 = 2.2 



and, therefore, that 
4-2 

Ay(2) = = 0.5, 



so yo(x 0 ) = 2, y 1 (x 0 ) = 2.5, yz(xo) = 3, y 3 (xo) = 3.5, y 4 (x 0 ) = 4, 



4.2-2.1 

Ay(2.1) = = 0.525, 

so y 0 (xi) = 2.1, yi(xi) = 2.625, y 2 (xi) = 3.15, y 3 (xj) = 3.675, 
y 4 Oi) = 4.2, 

4.4-2.2 

Ay(2.2) = = 0.55, 

so yo(x 2 ) = 2.2, yi(x 2 ) = 2.75, y 2 (x 2 ) = 3.3, yi(x 2 ) = 3.85, 
y 4 (x 2 ) = 4.4. 

(See Figure 5.128.) Thus, 

T 2A = ( °" 1) 4 (Q,5) [(2 3 + 2 2 ) + 2 ((2 3 + 2.5 2 ) + (2 3 + 3 2 ) + (2 3 + 3.5 2 )) 
+ (2 3 + 4 2 )] 

+ (0,1) ^' 525) [2(2. 1 3 + 2.1 2 ) + 4(2. 1 3 + 2.625 2 ) + 4(2. 1 3 + 3.15 2 ) 

+ 4(2. 1 3 + 3.675 2 ) + 2(2. 1 3 + 4.2 2 )] 
+ (0 - 1 ^ 0 - 55 ^ [(2.2 3 + 2.2 2 ) + 2(2.2 3 + 2.75 2 ) + 2(2.2 3 + 3.3 2 ) 

+ 2(2.2 3 + 3.85 2 ) + 2(2.2 3 + 4.4 2 )] 
= 0.0125(139) + 0.02625(156.7755) + 0.01375(175.934) = 8.271949375. 



5.7 | Numerical Approximations of Multiple Integrals (optional) 



In this case, the exact answer is 

.2.2 plx p2.2 



p2.2 p2x n2.2 n2.2 

J J (x 3 + .v 2 ) dy dx = J (x 3 y + ±y 3 ) |^ dx = J (x 4 + |x 3 ) dx 



= 8.23886. 



I 2 - 2 _ 1/T)5 
12 5 



(2.2 5 - 2 5 ) + i (2.2 4 - 2 4 ) 



If Z) is an elementary region of type 2, that is, 

D = {(x, y) e R 2 | a(y) < x < fi(y), c < y < d}, 

then we may similarly approximate ff D f dA by first partitioning the y-interval 
[c, d] using 

d — c 

Ay = , c = y 0 < yi < . . . < y„ = d, where y ; = c + jAy, 

n 

and then, for a fixed y in [c, d], by partitioning the corresponding x-interval 
[a(y), /5(y)] into m equal subintervals: 

£00 - 



Ajc(y) = 



77? 



a(y) = x 0 < Xi < . . . < x m = /3(y), where x,(y) = a(y) + i Ax(y). 
In doing so, we obtain a counterpart formula to that of (8), namely, 



T 

± in 



Ax(c)Ay 



4 

(-1 



| g Ax(y y -)Ay 



/(a(c), c) + 2 /(*/(c), c) + /(/3(c), c) 

m-l 

2/(«(y,).yD • 4j]/(.v,(y, ). y ; )) 



7 = 1 



1=1 



+ 2f(P(yj),yj) 



(9) 



+ 



Ax (J) Ay 



m-l 



/(a(d), d) + 2j2 f(xi(d), d) + f (J3(d), d) 



i=i 



Of course, we may also adapt Simpson's rule for use in the case of double 
integrals over elementary regions. Moreover, we may derive similar methods for 
approximating triple integrals as well. In practice, however, other methods are 
often used that lend themselves to computer implementation. 

One such alternative technique is known as the Monte Carlo method. It is 
based on a result called the mean value theorem for double integrals: If / is 
a continuous function of two variables and D is a bounded and connected (i.e., 
one-piece) region in R 2 , then there is a point P e D such that 



f(x, y)dA = f(P) ■ area of D. 



400 Chapter 5 | Multiple Integration 



5.7 Exercises 



Note that it follows that 

ffnf dA 
area ot D 

so that, by Definition 6.1, we have that f(P) = [f] w% , the average (mean) value 
of / on D. The Monte Carlo method for approximating ff D f dA is to select n 
points P\ , . . . , P n in D at random and compute the average value / of / on just 
these points: 

Then / will approximate [/] avg and, hence, 



- (area of D) 
fdA~ (area of D)f = K - T f(P,). 

n ' 4 



i=\ 



If the area of D is in turn difficult to determine exactly, it, too, may be estimated 
by situating D inside a rectangle R and selecting m points at random in R. If r 
denotes the fraction of these points lying in D, then area of D « r ■ (area of R). 

In an analogous manner, we may give Monte Carlo approximations for triple 
integrals of functions of three variables defined over solid regions in space. 



In Exercises 1—6, (a) use the trapezoidal rule approximation 
72,3 to estimate the values of the given integrals, and (b) com- 
pare your results with the exact answers. 

*3.1 /-2.1 



l.i r0.6 



pi. I nl.\ 

1. J J (x 2 -6y 2 ) dydx 

/■3.3 z-3.3 

2. / / xy 2 dy dx 



13 J3 
-2.2 fl.6 



dy dx 



n 

J [~Jx + <fy) dy dx 

/l.l /.0.6 
J e x+2y dydx 

6. / / x cos y dy dx 

J0 Jjt/6 

In Exercises 7—12, (a) use the approximation S2jf rom Simp- 
son s rule to estimate the values of the given integrals, and (b) 
compare your results with the exact answers. 
-OA /.0.3 



/U.l />U.J 
/ {y 4 - xy 2 ) dy dx 
-o.i Jo 

pOA p2 
JO Jl 



1 +X' 



dy dx 



Ii Jo 

Jo Jn 



? x+2}, dvdx 



-7i/4 rn/2 

10. / / sin2x cos3y dy dx 

rr/4 
px/4 rn/4 

11.1 / sin (x + y) dy dx 

Jo Jo 



1.1 rx/4 



Ii Jo 



e x cos v dy dx 



13. In Chapter 7 we will see that the area of the portion of 
the graph of f(x, y) for (x, y) in a region D in R 2 is 

given by the double integral / f D J f 2 + f 2 + 1 dA. 

(a) Set up an appropriate iterated integral to compute 
the surface area of the portion of the paraboloid 
z = 4 - x 2 - 3y 2 where (x, y) e [0, 1] x [0, 1]. 

(b) Use the trapezoidal rule approximation 74 4 to es- 
timate the surface area. 

14. Concerning the iterated integral 
f\' 5 f\.4 ln ( 2x + y) dydx: 

(a) Calculate the trapezoidal rale approximation T^. 

(b) Use Theorem 7.3 to estimate the error in your ap- 
proximation in part (a). 

(c) Calculate the Simpson's rule approximation 52,4. 



True/False Exercises for Chapter 5 401 



(d) Use Theorem 7.4 to estimate the error in your ap- 
proximation in part (a). 

15. Without either evaluating or estimating the integral 
fi 4 Jo s ln(xy)dy dx, which approximation is more 
accurate: 74 4 or 52,2? Explain your answer. 

1 6. Suppose that the trapezoidal rule is used to estimate the 
value of / 0 °' 2 /^qj e xl+2y dydx. Determine the small- 
est value of n so that the resulting approximation T„ t „ 
is accurate to within 10~ 4 of the actual value of the 
integral. 

17. Consider J 0 °' 3 J 0 °' 4 e x ~ y dy dx. 

(a) If the trapezoidal rule approximation T n „ is used 
to estimate the value of this integral, what is 
the smallest value of n so that the resulting 
approximation is accurate to within 10~ 5 of the 
actual value? 

(b) If the Simpson's rule approximation S2,,, i„ is used 
to estimate the value of this integral, what is the 
smallest value of n so that the resulting approx- 
imation is accurate to within 10~ 5 of the actual 
value? 

18. Concerning the iterated integral JqJq(3x + 
5y)dy dx: 

(a) Calculate the trapezoidal rule approximation 7/2,2. 

(b) Compare your result in part (a) with the exact 
answer. 

(c) Use Theorem 7.3 to explain your results in parts 
(a) and (b). 

19. Consider the iterated integral j\ f^ 2 x 3 y 3 dy dx: 

(a) Calculate the Simpson's rule approximation £2,2- 

(b) Compare your result in part (a) with the exact an- 
swer. 

(c) Use Theorem 7.4 to explain your results in parts 
(a) and (b). 

In Exercises 20-25, (a) use the approximation T3 3 from the 
trapezoidal rule to estimate the values of the given integrals. 

20. J J (x 3 + 2y 2 ) dy dx 



/*jr/4 pcosx 

21 . / / (2x cos y + sin 2 x) dy dx 

Jo J sinx 
,.0.3 fix 

22. J J (xy - x 2 ) dy dx 

23. / / dydx 
Jo Jo v 1 — y 2 

24. J j sin x dx dy (Note the order of integration.) 

/1.6 />2y 
J In (xy)dxdy 

26. In this problem, you will develop another way to think 
about the trapezoidal rule approximation given in equa- 
tion (6). 

(a) Let L denote a general linear function of two vari- 
ables, that is, L(x, y) = Ax + By + C, where A, 
B, and C are constants. Set R = [a, b] x [c, d]. 
Show that 

/ / LdA = (area of W)(average of the values of 
•> J R L taken at the four vertices of/?). 

(Note that this gives an exact expression for the 
double integral.) 

(b) Suppose that / is any function of two variables that 
is integrable on R. Show that the trapezoidal rule 
approximation Ti \ to ff R f dA is 

7^1 = (area of /?)(average of the values off 
taken at the four vertices of R). 

(c) Now let Ax = (b — a)/m, Ay = (d — c)/n, 
and, for i = 1 , . . . , m, j = 1, . . . , n, let /?,•_,• = 
[Xi-i, x{\ x [y,_i, y,], where x t = a + iAx and 
y/ = c + j Ay. Then we have 

///<" = ££ // f dA - 

Use 7i t i to approximate each integral ff R fdA 
and sum the results to obtain the formula for T„ un 
given by equation (6). 



True/False Exercises for Chapter 5 



1. Every rectangle in R 2 may be denoted [a, b] x [c, d]. 



f 1/2 f 2 3 2 

. I y sin {jix ) dy dx 



2. If / is a continuous function and f(x, y) > 0 on a 

region D in R 2 , then the volume of the solid in R 3 r 2 Z" 1 / 2 

under the graph of the surface z = f(x, y) and above = J i Jo ^ s ^ n ^ nx ^ dx 

the region D in the xy-plane is / / f(x,y)dA. 



402 Chapter 5 | Multiple Integration 



/ / 3 dy dx = J J 3 dx dy 

Jo Jo Jo Jo 



10 Jo 

-2 r y 



5- / / f(x,y)dxdy= I I f(x, y)dy dx for 
Jo Jo Jo Jo 

all continuous functions /. 

6. 0 < j J sin(x 4 + y 4 )dx dy < jt, where D denotes 
the unit disk {(x, y) \ x 1 + y 2 < 1}. 

7. f [ sfx 3 + ydydx = [ [ ^x 3 + y dx dy. 
Jo Jo Jo Jo 

8. J x 2 e x+y dydx=(^J x 2 e x dx\(j^ e y dy^j. 

9. The region in R 2 bounded by the graphs of y = x 3 
and y = ~Jx is a type 3 elementary region in the 
plane. 

10. The region in R 2 bounded by the graphs of y = sinx, 
y = cosx,x = u j A, andx = 5 jr/4 is a type 2 elemen- 
tary region in the plane. 

1 1 . The region in R 3 bounded by the graphs of y 2 — x 2 — 
z 2 = 1 and 9x 2 + 4z 2 = 36 is a type 3 elementary 
region in space. 

12. j j (y 3 + l)dxdy gives the area of the region D, 
where D = {(x, y) | (x - 2) 2 + 3y 2 < 5}. 

' // < 

angle with vertices (-1,0), (1, 0), (0, 1). 



13. / / (y 2 sin (x 3 ) + 3) dx dy = 3 /2, where D is the tri- 



(x + y 3 +z 5 )dV = 0. 

[-2,2]x[-l,l]x[-3,3] 

. [ [ I (x + z)dV = 0. 

/[-2,2]x[0,l]x[-l,l] 



SSL 
■SSL 

l l l[-2,2] 
Jo Jv/4 J- 



x[0,l]x[-l,l] 
4 n^y/2 f 2 



(y + z)dV = 0. 
sin yz dz dx dy = 0. 



18. / / / (y -x 2 )dzdydx = 0. 

J-l J-JT^x 2 J-^/l-x 2 -/ 

19. If T(n, v) = (2m — v, u + 3d), then the area of the im- 
age/) = T(D*)ofthe unit square/)* = [0, 1] x [0, 1] 
is 7 square units. 

20. If T(m, v) = (v — u,3u + 2v), then the area of the im- 
age D = T(D*) of the rectangle D* = [0, 3] x [0, 2] 
is 5 square units. 



21 



77 

JO Jy 



2 r5-2y 
12 



e 2x y cos(x — 3y) dx dy 



10 nu/2 



I f 

JO Ju 



-5e" cos vdvdu. 



22. If D is the disk {(x, y) | x 2 + y 2 < 9}, then 
jj v / 9-x 2 -y 2 dA = 36jr. 

/>2jt f>2 p4 

23. The iterated integral / / / dzdrdO represents 

Jo Jo J2r 

the volume enclosed by the cone of height 4 and 
radius 2. 



24. The iterated integral 



ffZS 

J-2 J-V^x 2 JO 



{2 + Jx 2 + y 2 )dzdy dx 



is given by an equivalent integral in cylindrical coor- 
dinates as 



2m p2 /.79-r 2 



pill pi r> 

Jo Jo Jo 



(r 2 + 2r)dz dr d6. 



10 Jo Jo 
25. The iterated integral 

fV2=^ n^4-x 2 -y 2 

/ / / 7x 2 + y 2 +z 2 +5dzdydx 

J-s/2.Jo J^/x 2 +y 2 



is given by an equivalent integral in spherical coordi- 
nates as 

/ / / Pv P 2 + 5 sin<pdpd<pd8. 
Jo Jo Jo 

26. The average value of the function /(x,y) = x 2 y 
over the semicircular region D = {(x, y) | 0 < y < 
V4 — x 2 } is given by 

1 rn/2 r2 



or by 



1 f' f 

- / / r 4 sin6»cos 2 6Uri/6» 
x Jo Jo 

— f [ r 4 sin9 cos 2 9 drd6. 

X Jn/2 Jo 



27. The center of mass of a lamina represented by the tri- 
angle with vertices (2, 0), (0, 1), (0, —1), and whose 
density varies as S(x, y) = (x 2 + l)cosy, has coordi- 
nates given by 



fo fu-2)/2 (* 3 + *) cos y d y dx 
r 2 rv-*)/i fY 2 



y = 0. 



Jo !(x-2)/2^ x2 + U cos y dy dx 

28. The centroid of a cone of radius a, height h, with axis 
the z-axis and vertex at (0, 0, h) is (0, 0, 

29. The center of mass of the solid cylinder of radius a, 
height h , with axis the z-axis whose density at any point 



Miscellaneous Exercises for Chapter 5 403 



30. The integral / / / p 5 sin2</5 siri(pdp dtp d8 repre- 
J J Jw 



varies as e d , where d is the square of the distance from 
the point to the z-axis is (0, 0, z), where 

f" f'' zre rl dzdrdO 

— \h. W with density z, expressed in spherical coordinates. 



r 2n r a r h .2 , , ,„ sents the moment of inertia about the z-axis of a solid 

Jo Jo Jo zre dzdrdd 



fo 2 " So So re * dzdr de 



Miscellaneous Exercises for Chapter 5 



1. Let B be the ball of radius 3; that is, 

B = {(x,y,z)\x 2 + y 2 +z 2 < 9}. 

Without resorting to any explicit calculation of an iter- 
ated integral, determine the value of the triple integral 
fff B (z 3 +2)dV by using geometry and symmetry 
considerations. 

2. Let W denote half of the solid ball of radius 2; that is, 

W = {(x, y, z) | x 1 + y 2 + z 2 < 4, z > 0}. 

Without resorting to explicit calculation of an iterated 
integral, determine the value of the triple integral 



SSL 



(x 3 +y-3)dV. 



(Hint: Use geometry and symmetry.) 

3. Let W be the solid region in R 3 with x > 0 that is 
bounded by the three surfaces z = 9 — x 2 , z = 2x 2 + 
y 2 , and x = 0. 

(a) Set up, but do not evaluate, two different (but 
equivalent) iterated integrals that both give the 
vzlueof fff w 3 dV. 

(b) Use a computer algebra system to find the value 
of Jff w 3dV and to check for consistency in your 
answers in part (a). 

4. Suppose that / is continuous on the rectangle R = 
[a, b] x [c, d]. For (x, y) G (a, b) x (c, d), we define 



F(x,y)- 



ff 

J a J c 



f(x', y')dy'dx' 



Use Fubini's theorem to show that d 2 F/dxdy = 
d 2 F/dydx. This provides an alternative proof of the 
equality of mixed partials. (Hint: Write 



where 



F(x, y) 



*(*', y) 



f g(x', 

J a 



y)dx' 



j" f(x',y')dy'. 



Then dF/dx and d 2 F/dydx may be calculated using 
the fundamental theorem of calculus. Then use Fubini's 
theorem to find dF/dy and d 2 F/dxdy.) 



5. Convert the following cylindrical integral to equivalent 
iterated integrals in (a) Cartesian coordinates and (b) 
spherical coordinates: 



p2n p\ p -\ 

Jo Jo Jo 



r dz dr d6. 

lo Jo Jo 

Evaluate the easiest of the three iterated integrals. 
6. The volume of a solid is given by the iterated integral 



Jo Jo J-^/i 



dz dx dy. 



(a) Sketch the solid and also describe it by giving equa- 
tions for the surfaces that form its boundary. 

(b) Express the volume as an iterated integral in cylin- 
drical coordinates. Determine the volume. 

7. Calculate the volume of a cube having edge length a 
by integrating in cylindrical coordinates. (Hint: Put the 
center of the cube at the origin.) 

8. Calculate the volume of a cube having edge length a 
by integrating in spherical coordinates. 



9. Determine 



cos 



2y 



X + V 



dA, 



where D is the triangular region bounded by the coor- 
dinate axes and the line x + y = 1 . 



10. Evaluate 



/>6 />1— 2y 
Jo J-2v 



y\x + 2y) 2 e (x+2y? dxdy 



by making a suitable change of variables. 

1 1 . Find the area enclosed by the ellipse E given by the 
equation 



~ + ~ b ~2 



1 



a 



in the following way: 

(a) First, write the area as the value of an appropriate 
iterated integral in Cartesian coordinates. Do not 
evaluate this integral. 

(b) Next, scale the variables by letting x = ax, 
y = by. To what region E* in the iy-plane does 



404 Chapter 5 , Multiple Integration 



the ellipse E correspond? Rewrite the xy-integral 
in part (a) as an xy-integral. 

(c) Finally, use polar coordinates to transform the xy- 
integral and thereby show that the area inside the 
original ellipse is nab. 

1 2. This problem concerns the rotated ellipse E with equa- 
tion ttx 1 + I4xy + 10y 2 = 9. 
(a) Let u = 2x — y, v = x + y and rewrite the equa- 
tion for E in the form 



+ 



b- 



l, 



where a and b are positive constants to be 
determined. 

(b) Use an appropriate change of variables and the re- 
sult of part (c) of Exercise 1 1 to find the area en- 
closed by E . 

13. Consider the ellipse E with equation 
5x 2 + 6xy + 5y 2 = 4. Let u = x — y, v = x + y and 
follow the steps of Exercise 12. 

1 4. Imitate the techniques of Exercise 1 1 to find the volume 
enclosed by the ellipsoid E given by the equation 



2 2 2 

x y Z 

h 1 

a 2 b 2 c 2 



1. 



15. Evaluate 



xy 



d y — x 



dA, 



where D is the region in the first quadrant bounded 



by the hyperbolas x 



y 



1, x — y = 4 and 



the ellipses x 2 /4 + y 2 = 1, x 2 /\6 + y 2 /4 = 1. (Hint: 
Sketch the region D, and use it to make an appropriate 
change of variables.) 



16. Evaluate 



(x 2 + y V 



dA, 



where D is the region in the first quadrant bounded 
by the hyperbolas x 2 — y 2 = 1, x 2 — y 2 = 9, xy = 1, 
xy = 4. 



17. Evaluate 



dA, 



where D is the region bounded by xy = 1, xy = 4, 

y = l, y = 2. 

1 8. (a) Generalizing the notions of double and triple inte- 
grals, develop a definition of the "quadruple inte- 
gral" // ff w fdVofa function f(x, y, z, w) over 
a four-dimensional region W in R 4 . 



mi 



(b) Use your definition in part (a) to calculate 
(x + 2y + 3z-4w)dV, 

r 

where W is the four-dimensional box 

W = {(x, y, z, w) | 0 < x < 2, -1 < y < 3, 
0 < z < 4, -2 < w < 2}. 

19. (a) Set up, but do not evaluate, a quadruple iterated in- 

tegral that computes the four-dimensional volume 
of the four-dimensional ball of radius a : 

B = {(x, y, z, w) | x 2 + y 2 + z 2 + w 2 < a 2 }. 

(b) Use a computer algebra system to give a formula 
for the volume. 

(c) Use a computer algebra system to give for- 
mulas for the n-dimensional volume of the 
M-dimensional ball 

B = {{x x ,x 2 , . . . , X n ) I x\ + x\ H Yx 2 n < a 2 } 

in the cases where n = 5, 6. Is there any pattern to 
your answers? 

In Exercises 20—23, you will give a general expression for the 
n-dimensional volume of the n-dimensional ball 

B = {(x x ,x 2 ,...,x n ) | x 2 +x 2 + ---x 2 < a 2 }. 

Let V„(a) denote this n-dimensional volume. Let C„ denote the 
n-dimensional volume of the unit ball U; that is, C„ = V„(l). 

20. By scaling the variables x\,x 2 , ■ ■ ■ ,x n and using a 
change of variables formula analogous to those in The- 
orems 5.3 and 5.5, show that V„(a) = C n a n . 

21. (a) Consider points in B of the form (jci, x 2 , 0, 0). 

Show that the coordinates x\, x 2 describe points 
(in R 2 ) lying in the disk of radius a. 

(b) In R", let the polar coordinates r and 9 replace 
the Cartesian coordinates X\ and x 2 . Argue that 
the points in B lying over a particular point (r, 6) 
in the disk described in part (a) must fill out an 
(n — 2)-dimensional ball of radius V 'a 2 — r 2 . 

(c) Use part (b) to show that the w-dimensional volume 
of B is given by 

f2jT 

■de. 



f I V n . 2 (Va 2 -r 2 )rdr, 
Jo Jo 



22. Use the previous two exercises to establish the recur- 
sive formula 

(2jta 2 \ 
V n (a) = V n - 2 {a). 



23. (a) Show that V\{a) = 2a and V 2 (a) = it a 2 . (These 
are familiar facts.) 



Miscellaneous Exercises for Chapter 5 405 



(b) Use part (a) and the previous problem to show that 



V n (a) : 



7T"' 2 

(n/2)! 

2(n+l)/2 jr («-l)/2 



if n is even 
a" if n is odd 



where the double factorial n\\ = n(n — 2)(« — 
4) • • • 3 • 1 (i.e., the product of all odd integers 
from 1 to n). 

24. A spherical shell with inner radius 3 cm and outer 
radius 4 cm has a mass density that varies as 
0.12c? 2 g/cm 3 , where d denotes the distance (in cen- 
timeters) from a point in the shell to the center of the 
shell. 

(a) Determine the total mass of the shell. 

(b) Will the shell float in water? (Note: The density 
of water is 1 g/cm 3 . To answer this question, you 
need to determine the average density of the shell.) 

(c) Suppose that the shell has a small hole so that the 
core of the shell fills with water. Now will it float? 

25. A dome is shaped as a hemisphere. If a pole whose 
length is the average height of the dome is to be in- 
stalled inside the dome in an upright position, where 
on the floor can it be located? 

26. Let / be continuous on R = [a, b] x [c, d]. In this 
problem, you will establish Leibniz's rule for "differ- 
entiating under the integral sign": 



d 

dy 



f 



f(x, y)dx 



J a 



(x, y) dx. 



(a) Let G(y') = f y (x, y')dx. For c < y < d, use 
the fundamental theorem of calculus to compute 
d/dy P G(y')dy' and, therefore, 



d 
dy 



- / / f y {x,y')dxdy' . 

y Jc J a 



(b) Use Fubini's theorem and part (a) to establish 
Leibniz's rule. 

27. The function f(x, y) = 1/ ^fxy is unbounded when ei- 
ther x or y is zero. Thus, if D = [0, 1] x [0, 1], we 

say that j j — dA is an improper double inte- 



>xy 

gral, analogous to the one-variable improper integral 
f 1 1 

/ — — dx. Improper multiple integrals of this type 
Jo V* 

may be evaluated using an appropriate limiting pro- 
cess. In this problem, you will determine the value of 

j j — d A in the following manner: 



Id Jxy 



(a) For 0 < e < 1/2, 0<<5<l/2, let D,j = 
[e, 1 - e] x [8, 1 - 8]. (Note that D € . s c D.) 
Calculate 



7(6, 8) : 



1 



dA. 



(b) Evaluate lim I(e, 8). You should obtain a fi- 

M)-K0,0) 

nite value, which may be taken to be the value of 
j j — dA, since D ( s "fills out" D as (e, 8) 



Id -jxy 

(0, 0). (We say that in this case the improper inte- 
gral converges.) 

28. Imitate the techniques of Exercise 27 to determine if 
the improper double integral 



a 



i 



dA 



[o,i]x[o,i] x + y 

converges and, if it does, find its value. (Hint: You will 
need to determine lim„_ ! .o+ u In u.) 

29. Imitate the techniques of Exercise 27 to determine if 
the improper double integral 



I I\o, 



dA 



/[o,i]x[o,i] y 

converges and, if it does, find its value. 

30. Calculate / f D In y ' x 1 + y 2 dA, where D is the unit 

disk x 2 + y 2 < 1 . Note that the integrand is not de- 
fined at the origin, so this is an example of an improper 
double integral. Nonetheless, you can find its value by 
integrating over the annular region 



{(x,y) | € <x 2 + y 2 < 1} 



and taking appropriate limits. 

31 . Find the value of the improper triple integral 

In ^x 2 + y 2 + z 2 dV, 

where B is the solid ball x 2 + y 2 + z 2 < 1 . (See Exer- 
cise 30.) 

32. If D is an unbounded region in R 2 , the integral 
ff D f(x, y)dA is another type of improper double 
integral, analogous to one-variable improper integrals 
such as / fl °° f(x)dx, f(x)dx, or f™^ f(x) dx. In 
this problem, you will determine the value of 



IL 



i 



d x 2 y 3 



dA, 



where D = {(x, y) | x > 1, y > 1} using a limiting 
process. 



406 Chapter 5 | Multiple Integration 



(a) For a > 1, b > 1, let D a j, = [1, a] x [1, b]. Com- 



pute 



I(a,b) 



dA. 



(b) Evaluate lim I (a, b). You should obtain a 

(a,b)— >(oo,oo) 

finite value, which may be taken to be the value 

of f f — r — - dA. (In such a case, we say that the 

J Jd * 2 r 
improper integral converges.) 

33. Let D = l(x, y) | x > 1, y > 1}. For what values of p 
and q does the improper integral 

1 



11 



xPyi 



dA 



converge? For those values of p and q for which the 
integral converges, what is the value of the integral? 

34. This problem concerns the improper integral 

n + x 2 + y 2 y ] dA, 



R 

where p is a constant. 

(a) Determine if the integral converges when p = —2 
by integrating over the disk D a = {(x, y) \ x 2 + 
y 2 < a 2 } and then letting a — > oo. 

(b) Determine for what values of p the integral 
//jj2(1 + x 2 + y 2 ) p dA converges. What is the 
value of the integral when it converges? 



35. Determine if 



SSL 



1 



R 3 (1 + x 2 + y 2 + z 2 fl 2 



dV 



converges by integrating over the ball B a = [(x, y, z) 
x 2 + y 2 + z 2 < a 2 } and then letting a — > oo. 



36. Show that 



SSL 



-Jx^+Z 1 d y 



converges by determining its value. 

37. In this problem, you will find the value of the one- 
variable improper integral f™ e~ x dx by using two- 
variable improper integrals. 

(a) First argue that J°° e~ xl dx converges. (Hint: 

Note that e~ xl < l/x 2 for all x; compare 
integrals.) 

(b) Let I denote the value of e~ xl dx. Show that 



/OO rC 
-oo J — c 



e x y dx dy 



/r 2 

(c) Let D a denote the disk x 2 + y 2 < a 2 . Evaluate 

ff Da e-* 2 -y>dA. 



SL<-' 



dA. 



(d) Compute I 2 as lima^oo 

(e) Now find e~ xl dx. 



'Da 



dA. 



Exercises 38-46 involve the notion of probability densities. A 
probability density function of a single variable is any func- 
tion f(x)such that f(x) > 0 for all x G R, and f(x) = 1. 
Given such a density function, the probability that a randomly 
selected number x falls between the values a and b is 



Prob(a < x < b) 



f 

J a 



f{x)dx. 



38. (a) Check that f(x) = e 2 '*' is a probability density 
function. 

(b) Egbert turns on the stove in a random manner to 
heat cooking oil to fry chicken. If the probability 
density of the temperature x of the oil is given by 
f(x) = i e -l x - 30 °l, what is the probability that the 
oil has a temperature between 250 °F and 350 °F? 

A joint probability density function for two random variables 
x and y is a function fix, y) such that 

(i) f(x, y) > 0 for all (x, y) e R 2 , and 

(ii) // R 2 /(*, y) dA = JZo f-oo y)dx dy=\. 
If f is such a probability density and D is a region in R 2 , then 
the probability that a randomly chosen point (x, y) lies in D is 



Prob((jc, y) e D) = 

39. (a) Show that the function 

I2x + y 
140 
0 



f(x,y)dA. 



if 0 < x < 5, 0 < v < 4 



otherwise 
is a joint probability density function, 
(b) Find the probability that x < 1 and y < 1 . 
40. (a) Show that the function 



f(^y) 



ye 



0 



if x > 0, y > 0 
otherwise 



is a joint probability density function, 
(b) What is the probability that x + y < 2? 

41 . If a and b are fixed positive constants, what value of C 
will make the function f(x, y) = Ce~ a ^~ h ^ a joint 
probability density function? 

42. Let a and b be fixed nonnegative constants, not both 
zero. For what value of C is 



fi^y) 



\C(ax + by) if 0 < x < 1, 0 < y < 1 
0 otherwise 
a joint probability density function? 



Miscellaneous Exercises for Chapter 5 407 



43. Let a and b be fixed positive constants and, for a given 
constant C, consider the function 



f(x,y) 



Cxy if 0 < x < a, 0 < y <b 
0 otherwise 



(a) For what value of C is / a joint probability density 
function? 

(b) Using the value of C that you found in part (a), 
what is the probability that bx — ay > 01 

44. The research team for Vertigo Amusement Park deter- 
mines that the length of time x (in minutes) a customer 
spends waiting to participate in the new Drown Town 
water ride, and the length of time y actually spent in the 
ride, are jointly distributed according to the probability 
density function 

-j:/50-v/5 

fix, y) 



250" 



0 



if* > 0,y > 0 
otherwise 



Find the probability that a customer spends at most 
an hour involved with the ride (both waiting and 
participating). 

45. Suppose that you randomly shoot arrows at a circular 
target so that the distribution of your arrows is given 
by the probability density function 



fix, y) 



1 



where x and y are measured in feet. In the center of 
the target, there is a bull's-eye that measures 1 ft in 
diameter. (See Figure 5.129.) What is the probability 
of your hitting the bull's-eye? 




Figure 5.129 The circular 
target of Exercise 45. The 
shaded region is the 
bull's-eye. 

46. If x is a random variable with probability density func- 
tion f(x) and y is a random variable with probability 
density function g(y), then we say that x and y are 
independent random variables if their joint density 
function is the product of their individual density func- 
tions, that is, if 

F(x,y) = f(x)g(y). 



Suppose that an electrical circuit is designed with two 
identical components whose time x to failure (mea-